calculus success in 20 minutes a day2ndedition[1]

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CALCU

LUS

Successin

20M

inutesa

Day

MASTER CALCULUS IN JUST 20 MINUTES A DAY!

STUDY GUIDES/Mathematics

SKILL BUILDERS

� Packed with key calculus concepts including ratesof change, optimization, antidifferentiation,techniques of integration, and much more

� Includes hundreds of practice questionswith detailed answer explanations

� Measure your progress with pre–and posttests

� Build essential calculus skills for successon the AP exams!

SKILL BUILDERS

CALCULUS ESSENTIALS INSIDE:

• Functions • Trigonometry • Graphs • Limits • Rates of change • Derivatives• Basic rules • Derivatives of sin(x) and cos(x) • Product and quotient rules • Chain

rule • Implicit differentiation • Related rates • Graph sketching • Optimization• Antidifferentiation • Areas between curves • The fundamental theorem of

calculus • Techniques of integration • and more!

CALCULUS SUCCESS PRACTICE

A good knowledge of calculus is essential for success on many tests and applicable fora wide range of careers. Calculus Success in 20 Minutes a Day helps students refresh andacquire important calculus skills. This guide provides a thorough review that fits into anybusy schedule. Each step takes just 20 minutes a day!

Pretest—Pinpoint your strengths and weaknesses

Lessons—Master calculus essentials with hundreds of exercises

Posttest—Evaluate the progress you’ve made

BONUS! Additional resources for preparing for important standardized tests ADDEDVALU

E—Access

toonline

practicew

ithInstantScoring

2ND EDITIONCompletely Revised and Updated!

FREE Calculus Practice!

Visit LearningExpress’s Online Practice Center to:

� Access additional calculus practice exercises

� Receive immediate scoring and detailed answer explanations

� Focus your study with our customized diagnostic report,and boost your overall score to guarantee success

Mark A. McKibben

CALCULUSSUCCESS

in 20 Minutes a Day

Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page i

Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page ii

CALCULUSSUCCESS

in 20 Minutes a Day

N E W Y O R K

®

Second Edition

Mark A. McKibbenChristopher Thomas

Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page iii

Copyright © 2012 LearningExpress, LLC.

All rights reserved under International and Pan-American Copyright Conventions.

Published in the United States by LearningExpress, LLC, New York.

Library of Congress Cataloging-in-Publication Data:

McKibben, Mark A.

Calculus success in 20 minutes a day / Mark A. McKibben.—2nd ed.

p. cm.

Previous ed.: Calculus success in 20 minutes a day / Thomas, Christopher. © 2006.

ISBN 978-1-57685-889-9

1. Calculus—Problems, exercises, etc. I. Thomas, Christopher, 1973– Calculus success

in 20 minutes a day. II. Title. III. Title: Calculus success in twenty minutes a day.

QA303.2.T47 2012

515—dc23

2011030506

Printed in the United States of America

9 8 7 6 5 4 3 2 1

ISBN 978-1-57685-889-9

For information or to place an order, contact LearningExpress at:

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Or visit us at:

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Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page iv

Mark A. McKibben is a professor of mathematics and computer science at Goucher College in Baltimore,

Maryland. During his 12 years at this institution, he has taught more than 30 different courses spanning the

mathematics curriculum, and has published two graduate-level books with CRC Press, more than two dozen

journal articles on differential equations, and more than 20 supplements for undergraduate texts on algebra,

trigonometry, statistics, and calculus.

Christopher Thomas is a professor of mathematics at the Massachusetts College of Liberal Arts. He has

taught at Tufts University as a graduate student, Texas A&M University as a postdoctorate professor, and the

Senior Secondary School of Mozano, Ghana, as a Peace Corps volunteer. His classroom assistant is a small

teddy bear named ex.

ABOUT THE AUTHOR

v

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Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page vi

INTRODUCTION ix

PRETEST 1

LESSON 1 Functions 15

LESSON 2 Graphs 23

LESSON 3 Exponents and Logarithms 31

LESSON 4 Trigonometry 37

LESSON 5 Limits and Continuity 47

LESSON 6 Derivatives 55

LESSON 7 Basic Rules of Differentiation 61

LESSON 8 Rates of Change 67

LESSON 9 The Product and Quotient Rules 75

LESSON 10 Chain Rule 81

LESSON 11 Implicit Differentiation 85

LESSON 12 Related Rates 91

LESSON 13 Limits at Infinity 97

LESSON 14 Using Calculus to Graph 107

LESSON 15 Optimization 115

CONTENTS

vii

Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page vii

LESSON 16 The Integral and Areas under Curves 121

LESSON 17 The Fundamental Theorem of Calculus 127

LESSON 18 Antidifferentiation 133

LESSON 19 Integration by Substitution 139

LESSON 20 Integration by Parts 145

POSTTEST 151

SOLUTION KEY 165

GLOSSARY 193

ADDITIONAL ONLINE PRACTICE 197

–CONTENTS–

vii i

Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page viii

INTRODUCTION

ix

If you have never taken a calculus course, and now find that you need to know calculus—this is the book

for you. If you have already taken a calculus course, but felt like you never understood what the teacher

was trying to tell you—this book can teach you what you need to know. If it has been a while since you

have taken a calculus course, and you need to refresh your skills—this book will review the basics and reteach

you the skills you may have forgotten. Whatever your reason for needing to know calculus, Calculus Success

in 20 Minutes a Day will teach you what you need to know.

Overcoming Math Anxiety

Do you like math or do you find math an unpleasant experience? It is human nature for people to like what

they are good at. Generally, people who dislike math have not had much success with math.

If you have struggles with math, ask yourself why. Was it because the class went too fast? Did you have

a chance to fully understand a concept before you went on to a new one? One of the comments students fre-

quently make is, “I was just starting to understand, and then the teacher went on to something new.” That is

why Calculus Success is self-paced. You work at your own pace. You go on to a new concept only when you

are ready.

When you study the lessons in this book, the only person you have to answer to is you. You don’t have

to pretend you know something when you don’t truly understand. You get to take the time you need to under-

stand everything before you go on to the next lesson. You have truly learned something only when you

thoroughly understand it. Take as much time as you need to understand examples. Check your work with the

answers and if you don’t feel confident that you fully understand the lesson, do it again. You might think you

don’t want to take the time to go back over something again; however, making sure you understand a lesson

Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page ix

completely may save you time in the future lessons.

Rework problems you missed to make sure you don’t

make the same mistakes again.

How to Use This Book

Calculus Success teaches basic calculus concepts in 20

self-paced lessons. The book includes a pretest, a

posttest, 20 lessons, each covering a new topic, and a

glossary. Before you begin Lesson 1, take the pretest.

The pretest will assess your current calculus abilities.

You’ll find the answer key at the end of the pretest.

Each answer includes the lesson number that the

problem is testing. This will be helpful in determin-

ing your strengths and weaknesses. After taking the

pretest, move on to Lesson 1, Functions.

Each lesson offers detailed explanations of a

new concept. There are numerous examples with

step-by-step solutions. As you proceed through a les-

son, you will find tips and shortcuts that will help

you learn a concept. Each new concept is followed by

a practice set of problems. The answers to the prac-

tice problems are in an answer key located at the end

of the book.

When you have completed all 20 lessons, take

the posttest. The posttest has the same format as the

pretest, but the questions are different. Compare the

results of the posttest with the results of the pretest

you took before you began Lesson 1. What are your

strengths? Do you still have weak areas? Do you need

to spend more time on some concepts, or are you

ready to go to the next level?

Make a Commitment

Success does not come without effort. If you truly

want to be successful, make a commitment to spend

the time you need to improve your calculus skills.

So sharpen your pencil and get ready to begin

the pretest!

–INTRODUCTION–

x

Calc2e_00_i-x_FM.qxd 11/18/11 12:32 AM Page x

Before you begin Lesson 1, you may want to get an idea of what you know and what you need to learn.

The pretest will answer some of these questions for you. The pretest consists of 50 multiple-choice

questions covering the topics in this book. While 50 questions can’t cover every concept or skill taught

in this book, your performance on the pretest will give you a good indication of your strengths and

weaknesses.

If you score high on the pretest, you have a good foundation and should be able to work through the

book quickly. If you score low on the pretest, don’t despair. This book will explain the key calculus concepts,

step by step. If you get a low score, you may need to take more than 20 minutes a day to work through a les-

son. However, this is a self-paced program, so you can spend as much time on a lesson as you need. You decide

when you fully comprehend the lesson and are ready to go on to the next one.

Take as much time as you need to complete the pretest. When you are finished, check your answers with

the answer key at the end of the pretest. Along with each answer is a number that tells you which lesson of

this book teaches you about the calculus skills needed to answer that question. You will find the level of dif-

ficulty increases as you work your way through the pretest.

PRETEST

1

Calc2e_00_1-14_Pre.qxd 11/18/11 12:34 AM Page 1

3

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

1.2.3.4.5.6.7.8.9.

10.11.12.13.14.15.16.17.

18.19.20.21.22.23.24.25.26.27.28.29.30.31.32.33.34.

35.36.37.38.39.40.41.42.43.44.45.46.47.48.49.50.

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

–LEARNINGEXPRESS ANSWER SHEET–


1. What is the value of f(4) when f(x) = 3x2 – ?

a. 22

b. 46

c. 140

d. 142

2. Simplify g(x + 3) when g(x) = x2 – 2x + 1.

a. x2 + 2x + 4

b. x2 – 2x + 4

c. x2 + 4x + 16

d. x2 + 4x + 13

3. What is (f ° g)(x) when f(x) = x – and g(x) =

x + 3?

a.

b.

c.

d.

4. What is the domain of ?

a. all real numbers except x = 1

b. all real numbers except x = 0

c. all real numbers except x = –1 and x = 1

d. all real numbers except x = –1, x = 0, and

x = 1

Use the following figure for questions 5 and 6.

5. On what interval(s) is f(x) increasing?

a. (�∞,1) and (5,∞)

b. (1,5)

c. (1,6)

d. (5,∞)

6. Which of the following is a point of inflection

for f(x)?

a. (0,5.5)

b. (1,6)

c. (3,3)

d. (5,1)

7. What is the equation of the straight line pass-

ing through (2,5) and (�1,�1)?

a.

b.

c.

d. y � �2x � 3

y � �2x � 9

y � 2x � 1

y � 2x � 5

h1x 2 �x

x2 � 1

x � 3 �2

x � 3

x2 � 2 � 3x �6x

2x �2x

� 3

x �2x

� 3

2x

x

–PRETEST–

5

1

2

3

4

5

6

–1–2–3 1 2 3 4 5 6–1

y

x

y = f(x)


6

8. Simplify .

a. 4

b. 8

c. 32

d. 4,096

9. Simplify .

a.

b. 8

c. �8

d. �6

10. Solve for x when 3x = 15.

a. 5

b.

c.

d.

11. Evaluate sin .

a.

b.

c.

d.

12. Evaluate .

a. �1

b. 1

c.

d.

13. Evaluate .

a. �1

b.

c.

d.

14. Evaluate .

a. 0

b. 1

c.

d. undefined

15. Evaluate .

a. ∞b. �∞

c.

d.−3

2

�14

limxS2�

x � 3x � 2

12

limxS1

x � 1x2 � 1

79

1517

35

limxS4

x2 � 1x2 � 1

2

2

2

tan3

4

π

3

2

2

2

12

�12

π3

ln112 2

ln115 2

ln13 2

ln15 2

18

2�3

6412

–PRETEST–


16. What is the slope of at x � 5?

a. 2

b. 17

c. 3x

d. 3

17. What is the slope of at

x � 3?

a. 2

b. 8

c. 14

d.

18. Differentiate .

a.

b.

c.

d.

19. The height of a certain plant is H(t) =

inches after week. How fast is

it growing after two weeks?

a. 5 inches per week

b. 10 inches per week

c. 21 inches per week

d. 31 inches per week

20. What is the derivative of ?

a.

b.

c.

d.

21. Differentiate .

a.

b.

c.

d.

22. Differentiate

a.

b.

c.

d. 2xsin1x 2cos1x 2′ =g x( )

2xsin1x 2 � x2cos1x 2′ =g x( )

2x � cos1x 2′ =g x( )

2xcos1x 2′ =g x( )

g1x 2 � x2sin1x 2.

f ¿ 1x 2 �1x

� ex

f ¿ 1x 2 �1x

� ex

f ¿ 1x 2 � ln1x 2 � ex

f ¿ 1x 2 � ln1x 2 � ex

f 1x 2 � ln1x 2 � ex � 2

dy

dx� 2x � 3tan1x 2

dy

dx� 2x � 3cos11 2

dy

dx� 2x � 3sin1x 2

dy

dx� 2x � 3sin1x 2

y � x2 � 3cos1x 2

t � 1

41 �40t

12x2 � 5x �1x

′ =h x( )

12x2 � 5x′ =h x( )

12x2 � 5′ =h x( )

12x2′ =h x( )

h1x 2 � 4x3 � 5x � 1

2x � 2

g1x 2 � x2 � 2x � 1

f 1x 2 � 3x � 2

–PRETEST–

7


23. Differentiate .

a. 0

b.

c.

d.

24. Differentiate .

a.

b. –cot(x)

c.

d.

25. Differentiate .

a.

b.

c.

d.

26. Differentiate m(x) = .

a.

b.

c.

d.

27. Compute if .

a.

b.

c.

d.

28. Compute if .

a.

b.

c.

d.dy

dx� 8xsec1y 2

dy

dx� cos1y 2 � 8x

dy

dx� 8xcos1y 2

dy

dx� 8x � cos1y 2

sin1y 2 � 4x2dy

dx

dy

dx

x x

y= −3

2

2

dy

dx�

3x2

1 � 2y

dy

dx�

3x2 � y

2y � x

dy

dx� x2

y2 � xy � x3 � 5dy

dx

10x1x2 � 1 24′ =m x( )

51x2 � 1 24′ =m x( )

12x 25′ =m x( )

10x′ =m x( )

1x2 � 1 25

14x2 � 7 2e4x2�8′ =f x( )

8xe4x2�7′ =f x( )

e4x2�7′ =f x( )

e8x′ =f x( )

f 1x 2 � e4x2�7

sin1x 2cos1x 2dy

dx=

cos21x 2 � sin21x 2

cos21x 2dy

dx=

dy

dx=

sec21x 2dy

dx=

y � tan1x 2

ln1x 2 � 1

x2′ =j x( )

1 � ln1x 2

x2′ =j x( )

1x

′ =j x( )

′ =j x( )

ln1x 2

xj x( ) =

8

–PRETEST–


29. What is the slope of at ?

a. �1

b. 1

c. –

d.

30. If the radius of a circle is increasing at 4 feet

per second, how fast is the area increasing

when the radius is 10 feet?

a. 20p square feet per second

b. 80p square feet per second

c. 100p square feet per second

d. 400p square feet per second

31. The height of a triangle increases by 3 inches

every minute while its base decreases by 1 inch

every minute. How fast is the area changing

when the triangle has a height of 10 inches and

a base of 100 inches?

a. It is increasing at 145 square inches

per minute.

b. It is increasing at 500 square inches

per minute.

c. It is decreasing at 1,500 square inches

per minute.

d. It is decreasing at 3,000 square inches

per minute.

32. Evaluate .

a. 4

b. �4

c. 2

d. undefined

33. Evaluate .

a. �∞b. ∞c. �4

d. 4

34. Evaluate .

a.

b. 2

c. 3

d. 0

13

limln( )

x

x

x→−∞ +3 2

limx

x x

x x→−∞

+ ++ −

4 6 4

10 1

5

3

limx

x x

x→∞

− +−

4 5 2

1

2

2

3

3

3

3

1

2

3

2,

x2 � y2 � 1

–PRETEST–

9


35. Which of the following is the graph of

?

a.

b.

c.

d.

y �1

x � 2

–PRETEST–

10

1

2

3

–1–2 1 2 3 4–1

–2

– 3

y

x

1

2

3

–1–2 1 2 3 4–1

–2

–3

y

x

1

2

3

–1–2 1 2 3 4–1

–2

–3

y

x

1

2

3

–1–2 1 2 3 4–1

–2

–3

y

x


36. On what interval is con-

cave down?

a. (1,12)

b. (�6,5)

c.

d. (�1,1)

37. The surface area of a cube is increasing at a rate

of 3 square inches per minute. How fast is an

edge increasing at the instant when each side is

20 inches?

a. inch per minute

b. inch per minute

c. 80 inches per minute

d. 24,000 inches per minute

38. A box with a square bottom and no top must

contain 108 cubic inches. What dimensions

will minimize the surface area of the box?

a. 2 in. × 2 in. × 27 in.

b. 8 in. × 8 in. × 3 in.

c. 6 in. × 6 in. × 3 in.

d. 4 in. × 4 in. × 6.75 in.

39. If and , then

what is ?

a. �20

b. 1

c. 3

d. 9

40. What is ?

a. 2

b. 3

c. 10

d. 12

41. If is the area under the curve

between t � 0 and t � x, what is

?

a.

b.

c.

d. 0

42. Evaluate .

a.

b.

c.

d. x3 � 4x2 � 5x � c

x3 � 4x2 � 5x

6x � 8 � c

6x � 8

� 13x2 � 8x � 5 2 dx

14

x4 � 2x

3x2 � 4

x3 � 4x

g¿ 1x 2y � t3 � 4t

g1x 2

�4

0

f 1x 2 dx

�8

5

g1x 2 dx

�5

3

g1x 2 dx � �4�8

3

g1x 2 dx � 5

3

20

1

80

( , )− 3 3

g1x 2 � x4 � 6x2 � 5

–PRETEST–

11

1

2

3

4

1 2 3 4 5–1

y

x

(4,3)

y = f(x)


43. Evaluate .

a. 3

b. 9

c. 18

d.

44. Evaluate

a.

b.

c.

d.

45. Evaluate

a.

b.

c.

d.

46. Evaluate

a.

b.

c.

d.

47. Evaluate .

a.

b.

c.

d.

48. Evaluate .

a.

b.

c.

d.1

1252 6( )x c+ +

xx x c

23

6

2

1

32+

+

5 22 4( )x c+ +

1

622 6( )x c+ +

x x dx+( )2 52�

43

x2sin1x3 2 � c

43

x3sin1x3 2 � c

43

sin1x3 2 � c

4sin1x3 2 � c

�4x2cos1x3 2 dx

15

e5 � c

e5 � c

e5x � c

15

e5x � c

�e5x dx˛.

1

212ln x c− +

1

212ln( )x c− +

ln x c− +1

12x2

13x3 � x

� c

� xx2 � 1

dx˛.

�sin1x 2 � c

sin1x 2 � c

�cos1x 2 � c

cos1x 2 � c

�sin1x 2 dx .

81

2

�9

0

2x dx

–PRETEST–

12


49. Evaluate .

a.

b.

c.

d.

50. Evaluate

a.

b.

c.

d. xcos(x) – cos(x) + c

�12

x2cos1x 2 � c

12

x2cos1x 2 � c

�xcos1x 2 � sin1x 2 � c

�xsin1x 2 dx˛.

12

x2ln1x 2 �14

x2 � c

x2ln1x 2 �14

x2 � c

xln1x 2 � ln1x 2 � c

12

x2ln1x 2 � c

�xln1x 2 dx

–PRETEST–

13


–PRETEST–

14

Answers

1. b. Lesson 1

2. a. Lesson 1

3. d. Lesson 1

4. c. Lesson 1

5. a. Lesson 2

6. c. Lesson 2

7. b. Lesson 2

8. b. Lesson 3

9. a. Lesson 3

10. c. Lesson 3

11. d. Lesson 4

12. a. Lesson 4

13. c. Lesson 5

14. c. Lesson 5

15. b. Lesson 5

16. d. Lessons 6, 7

17. b. Lessons 6, 7

18. b. Lesson 7

19. b. Lesson 8

20. a. Lesson 8

21. d. Lesson 8

22. c. Lesson 9

23. c. Lessons 8, 9

24. a. Lesson 9

25. c. Lesson 10

26. d. Lesson 10

27. b. Lesson 11

28. d. Lessons 4, 11

29. c. Lesson 11

30. b. Lesson 12

31. a. Lesson 12

32. b. Lesson 13

33. b. Lesson 13

34. d. Lesson 13

35. a. Lesson 14

36. d. Lesson 14

37. a. Lesson 12

38. c. Lesson 16

39. d. Lesson 16

40. c. Lesson 16

41. a. Lesson 17

42. d. Lesson 18

43. c. Lesson 18

44. b. Lesson 18

45. d. Lesson 19

46. a. Lesson 19

47. b. Lesson 19

48. d. Lesson 19

49. d. Lesson 20

50. a. Lesson 20


Functions

A function is a way of matching up one set of numbers with another. The first set of numbers is called the

domain. For each of the numbers in the domain, the function assigns exactly one number from the other set,

the range.

LE

SS

ON

FUNCTIONS

Calculus is the study of change. It is often important to know when a quantity is increasing, when it

is decreasing, and when it hits a high or low point. Much of the business of finance depends on pre-

dicting the high and low points for prices. In science and engineering, it is often essential to know pre-

cisely how fast quantities such as temperature, size, and speed are changing. Calculus is the primary tool for

calculating such changes.

Numbers, which are the focus of arithmetic, do not change. The number 5 will always be 5. It never goes

up or down. Thus, we need to introduce a new sort of mathematical object, something that can change. These

objects, the centerpiece of calculus, are functions.

15

1

Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 15

For example, the domain of the function could

be the set of numbers {1, 4, 9, 25, 100}, and the range

could be {1, 2, 3, 5, 10}. Suppose the function takes 1

to 1, 4 to 2, 9 to 3, 25 to 5, and 100 to 10. This could be

illustrated by the following:

Because we sometimes use several functions in

the same discussion, it makes sense to give them

names. Let us call the function we just mentioned by

the name Eugene. Thus, we can ask, “Hey, what does

Eugene do with the number 4?” The answer is “Eugene

takes 4 to the number 2.”

Mathematicians like to write as little as possible.

Thus, instead of writing “Eugene takes 4 to the num-

ber 2,” we often write “Eugene(4) � 2” to mean the

same thing. Similarly, we like to use names that are as

short as possible, such as f (for function), g (for func-

tion when f is already being used), h, and so on. The

trigonometric functions in Lesson 4 all have three-

letter names like sin and cos, but even these are abbre-

viations. So let us save space and use f instead of

Eugene.

Because the domain is small, it is easy to write

out everything:

However, if the domain were large, this would get

very tedious. It is much easier to find a pattern and use

that pattern to describe the function. Our function f

just happens to take each number of its domain to the

square root of that number. Therefore, we can describe

f by saying:

f(a number) = the square root of that number

Of course, anyone with experience in algebra

knows that writing “a number” over and over is a waste

of time. Why not just pick a variable to represent the

number? Just as f is a typical name for a function, lit-

tle x is often used for a variable name. Using both, here

is a nice way to represent our function f :

f(x) =

This tells us that putting a number into the func-

tion f is the same as putting it into . Thus,

f(25) = = 5 and f(f) = = 2.

ExampleFind the value of g(3) if .g1x 2 � x2 � 2

425

x

f 11 2 � 1 f 14 2 � 2 f 19 2 � 3 f 125 2 � 5f 1100 2 � 10

1 S 1 4 S 2 9 S 3 25 S 5100 S 10

16

PARENTHESES HINT

It is true that in algebra, everyone is taught “parentheses mean multiplication.” This means that 5(2 + 7) = 5(9) = 45. If x is a variable, then x(2 + 7) = x(9) = 9x. However, if f is the name of a function, then f (2 + 7) = f (9) = the number to which f takes 9. The expression f (x) is pronounced “f of x” and not “f times x.” This cancertainly be confusing. But, as you gain experience, it will become second nature. Mathematicians use paren-theses to mean several different things and expect everyone to know the difference. Sorry!

Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 16

SolutionReplace each occurrence of x with 3.

g(3) � 32 � 2

Simplify.

g(3) � 9 � 2 � 11

ExampleFind the value of h(�4) if h(t) = t3 � 2t2 + 5.

SolutionReplace each occurrence of t with –4.

h(–4) = (–4)3 – 2(–4)2 + 5

Simplify.

h(–4) = –64 – 2(16) + 5 = –64 – 32 + 5 = –91

Practice

1. Find the value of when .

2. Find the value of when

.



5. Find the value of when m(t) = –5t 3.

6. Find the value of when

.

7. Suppose that after t seconds, a rock thrown off

a bridge has height

feet off the ground. What is the height above

the ground after 3 seconds?

8. Suppose that the profit on making and selling x

cookies is .

How much profit is made on selling 100 cookies?

Plugging Variables into Functions

Variables can be plugged into functions just as easily as

numbers can. Often, though, the result can’t be sim-

plified as much.

ExampleSimplify f(w) if f(x) = + 2x 2 + 2.

SolutionReplace each occurrence of x with w.

f(w) = + 2w 2 + 2

That is all we can say without knowing more about w.

ExampleSimplify if .

SolutionReplace each occurrence of t with (a � 5).

Multiply out and .

Simplify.

g1a � 5 2 � a2 � 7a � 11

g1a � 5 2 � a2 � 10a � 25 � 3a � 15 � 1

�31a � 5 21a � 5 22

g1a � 5 2 � 1a � 5 22 � 31a � 5 2 � 1

g1t 2 � t2 � 3t � 1g1a � 5 2

w

x

P xx x

( ),

dollars= − −2 10 000

102

s1t 2 � �16t2 � 20t � 100

h1x 2 � 2x � 23 x

h164 2

m −

1

5

f 1x 2 � 2f 17 2

h1t 2 � t2 �34

h a12b

g1x 2 � x3 � x2 � x � 1

g1�3 2

f 1x 2 � 2x � 1f 15 2

– FUNCTIONS–

17

(a + b)2 ≠ a2 + b2. Remember to FOIL (first, out-side, inside, last) to get (a + b)2 = a2 + 2ab + b2.

When multiplying, an even number of negativesresults in a positive number, whereas an odd num-ber of negatives results in a negative number.

Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 17

Example

Simplify if .

SolutionStart with what needs to be simplified.

Use to evaluate and .

Multiply out .

Cancel the and the .

Factor out an a.

Cancel an a from the top and bottom.

Practice

Simplify the following.

9. when

10. when

11.

12. g(x 2 + ) when

13.

14.

15. when h(x) = –2x + 1

16.

Composition of Functions

Now that we can plug anything into functions, we can

plug one function in as the input of another function.

This is called composition. The composition of func-

tion f with function g is written . This means to

plug g into f like this:

It may seem that f comes first in , read-

ing from left to right, but actually, the g is closer to the

x. This means that the function g acts on the x first.

ExampleIf f(x) = + 2x and , then what is the

composition ?

SolutionStart with the definition of composition.

(f ° g)(x) = f(g(x))

Use .

Replace each occurrence of x in f with .

Simplify.

( )( )f g x x xo = + + +4 7 8 14

( )( ) ( )f g x x xo = + + +4 7 2 4 7

4x � 7

( )( ) ( )f g x f xo = +4 7

g1x 2 � 4x � 7

( )( )f g xo

g1x 2 � 4x � 7x

( )( )f g xo

( )( ) ( ( ))f g x f g xo =

f � g

g x g xg x x

( ) ( )( )

+ − =2

23 when

h1x � a 2 � h1x 2

a

f x a f x

af x x

( ) ( )( )

+ − − + when = 2 5

g x g x g tt

t( ) ( ) ( )28

6− = − when

g1t 2 �8t

� 6tx

f x h f x

hf x

x

( ) ( )( )

+ − = when 1

2

f 1x 2 � x2 � 3x � 1f 1x � a 2

f 1x 2 � x2 � 3x � 1f 1y 2

2x � a

12x � a 2a

a

2xa � a2

a

�x2x2

x2 � 2xa � a2 � x2

a

1x � a 22

1x � a 22 � x2

a

f 1x 2f 1x � a 2f 1x 2 � x2

f 1x � a 2 � f 1x 2

a

f 1x 2 � x2f 1x � a 2 � f 1x 2

a

–FUNCTIONS–

18

Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 18

Conversely, to evaluate (g ° f )(x), we compute:

Use f(x) = + 2x.

Replace each occurrence of x in g with + 2x.

Simplify.

We can form the composition of more than two func-

tions. Just apply the functions, one at a time, working

your way from the one closest to x outward.

ExampleIf , g(x) = 2 – x, and h(x) = 4x, then

what is (f ° g ° h)(x)?

SolutionStart with the definition of composition.

(f ° g ° h)(x) = f (g(h(x)))

Use h(x) = 4x.

(f ° g ° h)(x) = f (g(4x))

Compute g(4x) by replacing each occurrence of x in g

with 4x.

g(4x) = 2 – 4x

Next, substitute this into the composition.

(f ° g ° h)(x) = f (g(4x)) = f (2 – 4x).

Replace every occurrence of x in f with 2 – 4x.

(f ° g ° h)(x) = f (2 – 4x) =

Simplify.

(f ° g ° h)(x) =

Practice

Using , , and h(x) = x

– , simplify the following compositions.

17. (f ° g)(x)

18. (g ° f )(x)

19. (f ° h)(t)

20. (f ° f )(z)

21. (h ° h)(w)

22. (g ° h)(16)

23. (h ° f ° g)(x)

24. (f ° h ° f )(2x)

Domains

When an expression is used to describe a function f (x),

it is convenient to think of the domain as the set of all

numbers that can be substituted into the expression

and get a meaningful output. This set is called the

domain. The range of the function is the set of all pos-

sible numbers produced by evaluating f at the numbers

in its domain.

In the beginning of the lesson, we considered the

function:

f(x) =

However, we left out a crucial piece of information: the

domain. The domain of this function consisted of only

the numbers 1, 4, 9, 25, and 100. Thus, we should have

written

f(x) = if x � 1, 4, 9, 25, or 100

Usually, the domain of a function is not given

explicitly like this. In such situations, it is assumed that

the domain is as large as it possibly can be, meaning that

x

x

x

g1x 2 � x3 � 2x2 � 1f 1x 2 �1x

3 41 8

−−

xx

( )( )2 4 1

2 2 4 3− +− −

xx

f xxx

( ) = +−

12 3

( )( )g f x x xo = + +4 8 7

( )( ) ( )g f x x xo = + +4 2 7

x

( )( ) ( )g f x g x xo = + 2

x

( )( ) ( ( ))g f x g f xo =

–FUNCTIONS–

19

This shows that the order in which you com-pute a composition matters! In general, ( f ° g)(x)≠ (g ° f )(x).

Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 19

it contains all real numbers that, when plugged into the

function, produce another real number. Specifically,

including a number in the domain cannot violate one

of the following two fundamental prohibitions:� Never divide by zero.� Never take an even root of a negative number.

ExampleWhat is the domain of ?

SolutionWe must never let the denominator be zero, so

x cannot be 2. Therefore, the domain of this function

consists of all real numbers except 2.

The prohibition against even roots (like square

roots) of negative numbers is less severe. An even root

of a negative number is an imaginary number. Useful

mathematics can be done with imaginary numbers.

However, for the sake of simplicity, we will avoid them

in this book.

ExampleWhat is the domain of g(x) = ?

SolutionThe numbers in the square root must not be negative,

so , thus . The domain consists

of all numbers greater than or equal to .

Do note that it is perfectly okay to take the square

root of zero, since = 0. It is only when numbers are

less than zero that even roots become imaginary.

ExampleFind the domain of k(x) = .

SolutionTo avoid dividing by zero, we need ,

so , thus and .

To avoid an even root of a negative number,

, so . Thus, the domain of k is

, , .

A nice way of representing certain collections of

real numbers is interval notation, as follows:

COLLECTION OF INTERVAL REAL NUMBERS NOTATION

a < x < b (a,b)

a ≤ x < b [a,b)

a < x ≤ b (a,b]

a ≤ x ≤ b [a,b]

x > a (a,∞)

x ≥ a [a,∞)

x < b (–∞,b)

x ≤ b (–∞,b]

All real numbers (–∞,∞)

Note: A parenthesis is used when we intend to NOT

include a point, whereas a square bracket is used when

we intend TO include a point.

The domain of the previous example would be

written as follows:

(–∞,–3), (–3,–2), and (–2,4]

Practice

Find the domain of each of the following functions.

Express your answers using interval notation.

25.

26. h(x) = x + 1

f xx x

( )( )( )

= −+ −

1

3 5 2

x � �2x � �3x � 4

x � 44 � x � 0

x � �2x � �31x � 3 2 1x � 2 2 � 0

x2 � 5x � 6 � 0

4

5 62

−+ +

x

x x

0

�23

x � �23

3x � 2 � 0

3 2x +

x � 2

f 1x 2 �3

x � 2

–FUNCTIONS–

20

Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 20

27.

28.

29.

30.

31.

32. g uu

u u( )

( )=

+ +8

3 4 3

k1x 2 �24 2 � x

x � 8

h1x 2 � 23 x

j zz z

z( )

( )( )= − ++

1 2

12

g1x 2 � x2 � 5x � 6

k1t 2 �12t � 5

–FUNCTIONS–

21

Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 21

Calc2e_01_15-22.qxd 11/18/11 12:35 AM Page 22

LE

SS

ON

GRAPHS

Afunction can be fully described by showing explicitly what happens at each number in its domain

(for example, 4 S 2) or by giving its formula (for example, f(x) = ). However, neither of these

provides a clear visual picture of the function.

Fortunately, René Descartes came up with the idea of a graph, a visual picture of a function. Rather than

say 4 S 2 or f(4) = 2, we plot the point (4,2) on the Cartesian plane, as in Figure 2.1.

x

2

23

2 up

4 over

x

y

1

2

3

4

(4,2)

1 2 3 4 5

Figure 2.1

Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 23

Practice

Plot the following points on a Cartesian plane.

1. (3,5)

2. (�3,4)

3. (2,�6)

4. (�1,�5)

5. (0,3)

6. (�5,0)

7. (0,0)

8.

For the function , plot the point

(x,f (x)) for the following values of x.

9. x � 3

10. x � 1

11. x � 0

12. x � �2

If we plotted the points (x,f (x)) for all x in the

domain of f(x) = (not just the whole numbers, but

all the fractions and decimals, too), then the points

would be so close together that they would form a con-

tinuous curve as in Figure 2.2.

The graph shows us several interesting charac-

teristics of the function f(x) = .

We can see that the function f(x) = is

increasing (the graph is going up from left to right)

and not decreasing (the graph is going down from left

to right).

The function f(x) = is concave down

because it bows downward (see Figure 2.3) like a frown

and not concave up like a smile (see Figure 2.4). We

report the input intervals in each case. So, we say that

f is increasing on (0,∞) and concave down on (0,∞).

ExampleAssume the domain of the function graphed in Figure

2.5 is all real numbers. Determine where the function

is increasing and decreasing, and where the function is

concave up and concave down.

x

x

x

x

f 1x 2 � x2 � 2x � 5

9

2

1

4,

24

NOTE ON FINDING COORDINATES

We put the y into the formula y = f (x) = to imply that the y-coordinates of our points are the numbers weget by plugging the x-coordinates into the function f .

x

1

2

1 2 3 4

y

x

xy = f (x) =

(0,0),

(1,1)

(4,2)

1—21—4

Figure 2.2

Figure 2.3

Figure 2.4

x

Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 24

SolutionThe function g is increasing up to the point at x � 2,

where it then decreases down to x � 8, and then

increases thereafter. Using interval notation, we say

that g increases on (�∞,2) and on (8,∞), and that g

decreases on (2,8).

The concavity of g is trickier to estimate. Clearly

g is concave down in the vicinity of x � 2 and concave

up starting around x � 7. The exact point where the

concavity changes is called a point of inflection. On this

graph, it seems to be at the point (5,4), though some

people might imagine it to actually be a bit on either

side. Thus, we say that g is concave down on (�∞,5)

and concave up on (5,∞).

Honestly, any information obtained by simply

eyeballing a graph is going to be a rough estimate. Is

the local maximum at (2,6), or is it at (2.0003,5.9998)?

There is no way to tell the difference without an actual

formula for f (x).

The point at (2,6) where g stops increasing and

begins to decrease is the highest point in its immedi-

ate vicinity and is called a local maximum. The point at

(8,3) is similarly a local minimum, the lowest point in

its neighborhood. These points tend to be of particu-

lar interest, especially in applications.

ExampleUse the graph of the function h(x) in Figure 2.6 to

identify the domain, where it is increasing and decreas-

ing, where it has local maxima and minima, where it is

concave up and down, and where it has points of

inflection.

25

1

2

3

4

5

6

–1–2 1 2 3 4 5 6–1

y

x

7

8

9 107 8

y = g (x)

Figure 2.5

MATHEMATICAL NOTATION NOTE

Out of context, an expression like (2,8) is ambiguous. Is this a single point with coordinates x � 2 and y � 8?Is this an interval consisting of all the real numbers between 2 and 8? Only the context can make clear whichis meant. If we read “at (2,8),” then this is a single point. If we read “on (2,8),” then it refers to an interval.

1

23

4

5

6

–1–2–3 1 2 3 4 5 6–1

–2

–3

–4

–5

–6

y

x–4

y = h(x)

Figure 2.6

Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 25

SolutionThe first thing to notice is that h has three breaks, or

discontinuities. If we wanted to trace the graph of h

with a continuous motion of a pencil, then we would

have to lift up the pencil at x � �2, x � 2, and at

x � 5. The little unshaded circle at (5,3) indicates a hole

in the graph where a single point has been taken out.

This means that x � 5 is not in the domain, just as x �

�2 has no point above or below it. The situation at x

� 2 is more interesting because x � 2 is in the domain,

with the point (the shaded-in circle) at (2,�2) repre-

senting h(2) � �2. All of the points immediately

before x � 2 have y-values close to y � 3, but then

there is an abrupt jump down to x � 2. Such jumps

occur often when describing real-life situations using

functions like the way the cost of postage leaps up as

soon as a letter weighs more than one ounce.

Because of the discontinuities, we must name

each interval separately, as in: h increases on (�∞,�2),

(�2,2), (2,5), and on (5,∞). As well, h is concave up on

(�∞,�2), (2,5), and on (5,∞), and concave down on

(�2,2).

There is a local minimum at (2,�2), because this

point there is the lowest in its immediate vicinity, say

for all 1 � x � 3. There is no local maximum in that

range because the y-values get really close to y � 3;

there is no highest point in the range because of the

unshaded circle.

Similarly, a point of inflection can be seen at x �

2 but not at x � �2 because there can’t be a point of

inflection where there is no point!

The line x = –2 is called a vertical asymptote

because the graph of f (x) begins to look more like this

line the closer the inputs get to –2. Because the graph

appears to flatten out like the straight horizontal line

y � 0 (the x-axis) as the graph goes off to the left, we

say that the graph of appears to have a hor-

izontal asymptote at y � 0. We will examine this more

closely in Lesson 13.

Practice

Use the graph of each function to determine the dis-

continuities, where the function is increasing and

decreasing, the local maximum and minimum points,

where the function is concave up and down, the points

of inflection, and the asymptotes.

y � h1x 2

–GRAPHS–

26

Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 26

13.

14.

15.

16.

–GRAPHS–

27

1

2

3

4

5

6

–1–2–3–4–5–6 1 2 3 4 5 6–1

–2

–3

–4

–5

–6

y

x

y = f (x)

1

2

3

–1

–2–3 1 2 3–1

y

x

4

y = g(x)

1

2

3

–1

–2–3 1 2 3–1

y

x

4

–2

y = h(x)

1

2

3

4

5

6

–1–2–3–4–5–6 1 2 3 4 5 6–1

–2

–3

–4

–5

–6

y

x

y = k(x)

Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 27

17.

18.

19.

20.

Note

We can obtain all sorts of useful information from a

graph, such as its maximal points, where it is increas-

ing and decreasing, and so on. Calculus will enable us

to get this information directly from the function. We

will then be able to draw graphs intelligently, without

having to calculate and plot thousands of points (the

method graphing calculators use).

–GRAPHS–

28

1

2

3

–1

–2–3 1 2 3–1

y

x

–2

–3

y = f(x)

(2,3)

1

2

–1

–2–3 1 2 3–1

y

x

–2

y = g(x)

1

2

3

4

5

6

–11 2 3 4 5 6–1

–2

–3

–4

y

x

y = j(x)

(2,5)

1

2

3

4

5

6

–1–2–3–4 1 2 3 4 5 6–1 7 8 9

y

x

y = h(x)

7

Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 28

Straight Lines

The most familiar and arguably most widely used of all

graphs are straight lines. Human beings tend to build

and move linearly. Given any two points, we can

immediately get a feel for the steepness of a line, as

seen in Figure 2.7.

“How much a line is increasing or decreasing” is

called the slope and is calculated by:

slope �

y-change�

x-change

�

ExampleWhat is the slope of the line passing through points

(2,7) and (�1,5)?

Solution

slope �

Practice

Find the slope between the following points.

21. (1,5) and (2,8)

22. (7,3) and (�2,3)

23. (�2,�4) and (�6,5)

24. (2,7) and (5,w)

Equation of a Line

No matter what two points you choose on a line, the

slope will always be the same. Thus, if a straight line

has slope m and goes through the point , then

using any other point (x,y) on the line, we get the same

slope, namely:

By cross-multiplying, we get the point-slope formula

for the equation of a straight line:

or equivalently

Going one step further, we get

y = mx +

This is called the slope-intercept formula for the line

because the point (0,b) is the y-intercept of the line

(that is, where it crosses the y-axis).

( )− +mx y

b

1 1

Call this 1 24 34

y � m1x � x1 2 � y1

y � y1 � m1x � x1 2

y � y1

x � x1� m

1x1, y1 2

5 � 7�1 � 2

��2�3

�23

y2 � y1

x2 � x1

riserun

–GRAPHS–

29

y

x

(x , y )1

(x , y )22

1

“rise”

y – y12

“run”x – x

12

Figure 2.7

Make sure to subtract the y’s in the top and the x’sin the bottom IN THE SAME ORDER. For instance,don’t use y2 – y1 for the rise and x1 – x2 for the run.

Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 29

ExampleFind the equation of the line with slope �2 through

point (�1,8). Graph the line.

Solution

(see Figure 2.8)

The slope of �2 � means the y-value goes

down 2 with every decrease of 1 unit in the x-value.

ExampleFind the equation of the straight line through (2,6)

and (5,7). Graph the line.

SolutionThe slope is , so the equation is

(see Figure 2.9).

The slope of means the y-value goes up 1 for

every increase of 3 units in the x-value.

Practice

Find the equation of the straight line with the given

information and then graph the line.

25. slope 2 through point (1,�2)

26. slope through point (6,1)

27. through points (5,3) and (�1,�3)

28. through points (2,5) and (6,5)

�23

13

y �13

1x � 2 2 � 6 �13

x �163

7 � 65 � 2

�13

�21

y � �2x � 6

y � �21x � 1�1 2 2 � 8

–GRAPHS–

30

1

2

3

4

5

6

–11 2 3 4–1

y

x

(0,6)

y = –2x + 6

(–1,8)

Figure 2.8

1

2

3

4

5

6

–1–2–3–4–5–6 1 2 3 4 5 6–1

y

x

0, y = x +

(2,6) (5,7)

16—3

16—31—3

Figure 2.9

Calc2e_02_23-30.qxd 11/18/11 12:37 AM Page 30

Exponents

Exponents frequently arise in calculations throughout calculus. If a is a positive real number and n is a pos-

itive integer (that is, n = 1, 2, 3, …), then an means “multiply the base a by itself n times.” Symbolically,

ExamplesReview the following examples.

25 � 2 # 2 # 2 # 2 # 2 � 32

34 � 3 # 3 # 3 # 3 � 81

n times

an � a # a # a p a

LE

SS

ON

EXPONENTS AND LOGARITHMS3

31

�

Do not multiply the base a times the exponent n.Symbolically, an ≠ a ⋅ n.

Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 31

106 = 1,000,000

When two numbers with the same base are multiplied,

their exponents are added.


The rule about adding exponents has an inter-

esting consequence. We know that

because this is what “square root” means. Alternately,

. Because and act

exactly the same, they are equal: � . This works

for square roots, cube roots, and so on:


When two numbers with the same base are divided,

their exponents are subtracted.

ExamplesWork through the following simplifications.

The rule about subtracting exponents has two inter-

esting consequences. First, � 1 because any

nonzero number divided by itself is one. Also,

. Thus, � 1. In general:

� 1


� 1

� 1

The second consequence follows from:

while

also . Thus, . In general:


When an is itself raised to a power m, the exponents

are multiplied.

51

5

1

2−

= = 1

5 1

2

4�1 �141 �

14

3�2 �132 �

19

a�n �1an

2�4 �124

23

27 � 23�7 � 2�4

23

27 �2 # 2 # 2

2 # 2 # 2 # 2 # 2 # 2 # 2�

12 # 2 # 2 # 2

�124

2000

30

a0

5054

54 � 54�4 � 50

54

54

11

1111 11

15

615 6 9= =−

35

32 �3 # 3 # 3 # 3 # 3

3 # 3�

3 # 3 # 3 # 3 # 3

3 # 3� 3 # 3 # 3 � 33

an

am � an�m

64 64 413 3= =

9 9 31

2 = =

a a a a a a1

213

143 4= = = , , , K

5125

51255

12 # 5

12 � 5

12 �1

2 � 51 � 5

5 5 5 ⋅ =

72 # 74 # 73 � 79

53 # 5 � 53 # 51 � 54

2

3

2

3

2

3

2

3

2

3

2

3

16

81

2 2

⋅

= ⋅ ⋅ ⋅ =

410 # 47 � 417

m timesn times

an # am � 1a # a # a p a 2 # 1a # a # a p a 2 � an�m

1

2

1

2

1

2

1

2

1

8

3

+ ⋅ ⋅ =

51 � 5

–EXPONENTS AND LOGARITHMS–

32

� �

Again, don’t multiply a times the exponent 0 toconclude that a0 = 0.

Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 32


Practice


1.

2.

3.

4.

5.

6.

7.

8. 5–1 ⋅ 5

9.

10.

11.

12.

13.

14.

Exponential Functions

We can form an exponential function by leaving the

base fixed and varying the exponent.

ExampleThe function has the graph shown in Fig-

ure 3.1. Note that is quite different from . For

example, when , the value of is

, while the

value of is .

ExampleThe function has the graph shown in Fig-

ure 3.2. Note that g (x) grows faster than f (x) = 2x as x

gets larger. For reasons that will become clear later, a

very nice base to use is the number e � 2.71828 . . . ,

which, just like p� 3.14159 . . . , can never be written

out completely.

g1x 2 � 3x

102 � 10 # 10 � 100x2

210 � 2 # 2 # 2 # 2 # 2 # 2 # 2 # 2 # 2 # 2 � 1,024

2xx � 10

x22x

f 1x 2 � 2x

9

81

12

2

2

−

−

4 4 44

2 0 5

1

−

−⋅ ⋅

8

8

4

2

12−( )

5 23−( )

2723

2�3

91

38 # 3 # 3�5

60

63

65

107

103

4 # 42

23 # 22

5 5 5 5 5

4 4 4 4 4 414

22

2 2 2 2 4

23

2 2 2 2 2 2 66

( ) = ⋅ = =

( ) = ⋅ ⋅ = = =

+

− − − − − + − + − −( ) ( ) ( )

a a a a a anm

n n n

m

n n n m

m

( ) = ⋅ ⋅ ⋅ = =+ + ⋅K1 244 344K

674 84

times

times


33

1

2

3

4

5

6

–1–2–3 1 2 3 4–1

y

x

(3,8)

(2,4)

(1,2)

(0,1)

xy = f(x) = 2

–1, 1—2–2, 1—4–3, 1—8

Figure 3.1

Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 33

Because 2 � e � 3, the graph of y � fits

between y � and y � (see Figure 3.3).

Another useful function is the inverse of ,

known as the natural logarithm ln(x). Just as subtract-

ing undoes adding, dividing undoes multiplying, and

taking a square root undoes squaring, the natural log-

arithm undoes .

If y � , then = x.

The graph of y � ln(x) comes from flipping the

graph of y � across the line y � x, as depicted in

Figure 3.4. In particular, since e0 = 1, it follows that

ln(1) = 0.

ex

ln1y 2 � ln1ex 2ex

ex

ex

3x2x

ex

34

1

2

3

4

5

6

–1–2 1 2 3–1

y

x

7

8

9 (2,3)

(1,3)

(0,1)

xy = g(x) = 3

–1, 1—3–2, 1—9

Figure 3.2

1

2

3

4

5

6

–1–2 1 2 3–1

y

x

7

8

9

(0,1)

x

x

y = 3

y = 2

y = e x

(1, e)

(1, e 2)

–2, 1—e2–1, 1—e

Figure 3.3

EXPONENTS AND LOGARITHMS

The exponential function f(x) = ex and the natural logarithm g(x) = ln(x) “undo” each other when composed.That is,

f (g(x)) = eln(x) = x and g(f(x)) = ln(ex) = x

We say that f and g are inverses.

ln(0) ≠ 1! In fact, ln is not even defined at x = 0.

Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 34

The laws of natural logarithms might appear unusual,

but they are natural consequences of the exponent rules.

The last of the three preceding laws is useful for

turning an exponent into a matter of multiplication.

ExampleSolve for x when .

SolutionTake the natural logarithm of both sides.

Use .

Divide both sides by ln(10).

A calculator can be used to find a decimal approxima-

tion: � 0.84509, if desired.

ExampleSimplify ln(25) � ln(4) � ln(2).

SolutionUse .

� ln(2)

Use .

� ln(50)

Practice


15.

16.

17.

18.

19. eln152

ln1e2 2

e0

e12

e5

e3 # e8

ln a25 # 4

2b

ln1a 2 � ln1b 2 � ln aabb

ln125 # 4 2

ln1a 2 � ln1b 2 � ln1a # b 2

ln17 2

ln110 2

x �ln17 2

ln110 2

x # ln110 2 � ln17 2

ln1an 2 � n # ln1a 2

ln110x 2 � ln17 2

10x � 7

ln1an 2 � n # ln1a 2

ln1a 2 � ln1b 2 � ln aabb

ln1a 2 � ln1b 2 � ln1a # b 2


35

1

2

3

–1

–2–3 1 2 3–1

–2

–3

y

x

(1,e)

(0,1) (e, 1)

(1,0)

y = e x

y = ln(x)

y = x

, –3 1—e3

, –2 1—e2

, –1 1—e

–3, 1—e3–2, 1—e2

–1, 1—e

Figure 3.4

These are the only three properties. Take particu-lar note of the following, which are often mistak-enly used in their place.ln(a + b) ≠ ln(a) + ln(b)ln(a – b) ≠ ln(a) – ln(b)(ln(a))b ≠ b ln(a)

Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 35

20.

21. ln(7) 1 ln(2)

22. ln(24) 2 ln(6)

23.

24. Solve for x when = 10.

25. Solve for x when = 100.

26. If loga x = 2 and loga y = –3, then what is

?loga

xy 3

3x # 35

2x

ln( ) ln( ) ln5 225

− +

ln 250( )


36

Calc2e_03_31-36.qxd 11/18/11 12:38 AM Page 36

LE

SS

ON

TRIGONOMETRY

Some very interesting and important functions are formed by dividing the length of one side of a right

triangle by the length of another side. These functions are called trigonometric because they come from

the geometry of a right triangle. Let H represent the length of the hypotenuse, A represent the length

of the side adjacent to the angle x, and O represent the length of the side opposite (away) from the angle x.

Such a triangle is depicted in Figure 4.1.

4

37

x

A

OH

Figure 4.1

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 37

The six trigonometric functions, sine (abbrevi-

ated sin), cosine (cos), tangent (tan), secant (sec), cose-

cant (csc), and cotangent (cot), are defined at an angle

x by dividing the following sides:

The first thing to notice is that all of the func-

tions can be obtained from just sin(x) and cos(x) using

the following trigonometric identities.

Thus, all of the trigonometric functions can be evalu-

ated for an angle x if the sin(x) and cos(x) are known.

The next thing to notice is that the Pythagorean

theorem, which, stated in terms of the sides O, A, and

H, is . And, if we divide through by ,

we get the following:

Thus, . To save on paren-

theses, we often write this as .

Because no particular value of x was used in the cal-

culations, this is true for every value of x.

Drawing triangles and measuring their sides is an

impractical and inaccurate method to calculate the

values of trigonometric functions. It is better to use a

calculator. However, when using a calculator, it is very

important to make sure that it is set to the same format

for measuring angles that you are already using: that is,

degrees or radians.

There are 360 degrees in a circle, possibly because

ancient peoples thought that there were 360 days in a

sin21x 2 � cos21x 2 � 1

1sin1x 2 22 � 1cos1x 2 22 � 1

aOHb

2

� aAHb

2

� 1

O2

H2 �A2

H2 �H2

H2

H2O2 � A2 � H2

cot1x 2 �AO

�A

H

OH

�cos1x 2

sin1x 2

csc1x 2 �HO

� 1 OH

�1

sin1x 2

sec1x 2 �HA

� 1 AH

�1

cos1x 2

tan1x 2 �OA

�O

H

AH

�sin1x 2

cos1x 2

cot1x 2 �AO

csc1x 2 �HO

sec1x 2 �HA

tan1x 2 �OA

cos1x 2 �AH

sin1x 2 �OH

38

MNEMONIC HINT

Some people remember the first three trigonometric functions by SOA CAH TOA to remember

sin(x) � , cos(x) � , and tan(x) � .OA

AH

OA

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 38

year. As the earth went around the sun, the position of

the sun against the background stars moved one degree

every day. The 2π radians in a circle correspond to the

distance around a circle of radius 1.

� To convert from degrees to radians, multiply by

.

� To convert from radians to degrees, multiply by

.

ExampleConvert 45° to radians.

Solution

radians � radians

Example

Convert radians to degrees.

Solution

radians �

Practice

Convert the following to radians.

1. 30°

2. 180°

3. 270°

4. 300°

5. 135°

Convert the following to degrees.

6.

7.

8. 2p

9.

10.11π

6

π10

π2

π3

2

3 120

ππ

⋅ ° = °1802π3

2π3

π4

45 45180

° = °°

⋅ π

3602

180° = °π π

2

360

π π°

=°180

39

CONVERSION HINT

To convert from degrees to radians, multiply by .

To convert from radians to degrees, multiply by .3602

180° = °π π

2360 180

π π°

=°

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 39

Trigonometric Values of Nice Angles

There are a few nice angles for which the trigonomet-

ric functions can be easily calculated. If x = = 45°,

then the two legs of the triangle are equal. If the

hypotenuse is H � 1, then we have the triangle in Fig-

ure 4.2.

By the Pythagorean theorem, , so

and . This means that O � A �

. If we rationalize the denominator, we

get . Thus:

Another nice angle is x = 60° = , because it is

found in equilateral triangles such as the one seen in

Figure 4.3. This triangle can be cut in half by inserting

a segment from the top vertex down to the base, result-

ing in the triangle shown in Figure 4.4.

π3

cosπ4 1

2

2

22

= = = A

H

sinπ4 1

2

2

22

= = = O

H

1

2

1

2

2

2

2

2 = =⋅

1

2

1

2 =

A2 �12

2A2 � 1

A2 � A2 � 1

π4

–TRIGONOMETRY–

40

O = A

A

π–4

H = 1

Figure 4.2

1 1

1

–π3

Figure 4.3

1

–π3

1–2

O

H =

A =

–π6

Figure 4.4

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 40

By the Pythagorean theorem, ,

so . Thus, . This

means that:

We can flip that last triangle around to calculate

the trigonometric functions for the other angle x = 30°

= (see Figure 4.5).

ExampleCompute sec .

SolutionUse the trigonometric identity for sec.

Use x = .

Use .

Simplify.

Practice

Use the trigonometric identities to evaluate the

following.

11. tan

12. tan

13. csc

14. sec

15. cot

16. cot

17. sec

18. cscπ4

π6

π6

π3

π3

π6

π3

π4

secπ4

2

2

2

2

2

2

2 22

2

= = ⋅ = =

secπ4 2

2

= 1

cosπ4 2

= 2

seccos( )

ππ4

1

4

=

π4

sec1x 2 �1

cos1x 2

π4

cosπ6 1

3

2

32

= = = A

H

sinπ6 1

1

2

12

= = = O

H

π6

cosπ3 1

1

2

12

= = = A

H

sinπ3 1

3

2

32

= = = O

H

O 3

4 = = 3

2O2 � 1 �

14

�34

a12b

2

� O2 � 12

–TRIGONOMETRY–

41

–π

1–2

3

2

6

A =

O =

H = 1

Figure 4.5

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 41

Trigonometric Values for Angles Greater Than 90° =

Example

For example, for , we have the picture shown

in Figure 4.6.

The circle of radius 1 around the origin is called

the unit circle. As such, the hypotenuse has length 1,

and the sine is the y-value of the point where a ray of

the given angle intersects with the circle of radius 1.

Similarly, the cosine is the x-value. Note in Figure 4.7

that the angle of measure 0 runs straight to the right

along the positive x-axis, and every other positive angle

is measured counterclockwise from there.

This can be used to find the trigonometric values

of nice angles greater than 90° = . The trick is to

use either a 30°, 60°, 90° triangle (a , , trian-

gle) or else a 45°, 45°, 90° triangle (a , ,

triangle) to find the y-value(sine) and x-value (cosine)

of the appropriate point on the unit circle. As before,

calculating the trigonometric values for non-nice

angles is best done with a calculator.

ExampleFind the sine and cosine of 120° = .

Solution

For this angle, we use a , , triangle, as

shown in Figure 4.8 to find the x- and y-values. The

y-value of the point where the ray of angle hits2

3

π

π2

π3

π6

2

3

π

π2

π4

π4

π2

π3

π6

π2

cosπ6

= 3

2

sinπ6

= 1

2

π6

30= °

π2

–TRIGONOMETRY–

42

1––2

32

,

x

y

1

–1

1

–1

–π6

–π630 =

–32

11–2

Figure 4.6

–

π

00° = 360° =

30° =

45° =

60° = 90° =

120° =

135° =

150° =

180° =

210° =

225° =

240° = 270° =

300° =

315° =

330° =

6

4

32

ππ

π

–

–

π

– 2π

3π

5π

7π

5π

4π3π

5π7π

11π

3

4

6

6

4

32

3

4

6

Figure 4.7

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 42

the unit circle is y � . Thus, .

The x-value is negative, x � , so .

ExampleFind the sine and cosine of = 225°.

SolutionBecause is a multiple of , we use a

triangle to find the x- and y-values. As

seen in Figure 4.9, both coordinates are negative, so

and .

ExampleFind all of the trigonometric values for 90° = .

Solution

Even though there isn’t a triangle here, there is still a

point on the unit circle. See Figure 4.10. We conclude

that cos = 0 and sin = 1 from the x- and

y-values of the point. Using the trigonometric identi-

ties, we can calculate that = 1 and

= 1. The tangent and secantcotcos( )

sin( )π π

π2

0

12

2

= =

cscsin( )

ππ2

1

2

=

π2

π2

π2

cos5

4

2

2

π

= − sin5

4

2

2

π

= −

45�, 45�, 90�

45�225�

5

4

π

cos2

3

1

2

π

= − �12

sin2

3

π

= 3

2

3

2

–TRIGONOMETRY–

43

120 =

–3π

1

2π–3

2π–3

32

1–232

,(– )

1–2–

Figure 4.8

225° =5π–4

5π–4

1

( )

4π–

2_2

2

2– 2–,

2_

2_ 2_

Figure 4.9

π90° =2–

(0,1)

Figure 4.10

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 43

functions, however, involve division by 0 and thus are

undefined. The angle x = is not in the domain of

tan and sec.

Notice that all of the trigonometric functions are

the same at 0° = 0 and 360° = 2π. This is because turn-

ing 360° leaves you facing in your original direction.

Thus, everything repeats at this point.

Using the table along with the fact that every-

thing repeats, we can sketch the graphs of and

. See Figures 4.11 and 4.12.

The functions sine and cosine are classic exam-

ples of periodic, or oscillating, functions.

cos1x 2

sin1x 2π2

–TRIGONOMETRY–

44

1

– π6 4 3 2

ππ π–– – 2π 3π 5π π π π π π π π7 5 4 3 5 73 4 6 6 4 3 2 3 4

π π π π π π116

232

–– – –– –4

––6

1–2

1–2–

22

22

32

32

13 96 4

73

52

y = sin(x)

ππ ππ

––

–1

Figure 4.11

1

– π6 4 3 2

ππ π–– – 2π 3π 5π π π π π π π π7 5 4 3 5 73 4 6 6 4 3 2 3 4

π π π π π π116

232

–– – –– –4

––6

1–2

1–2–

22

22

32

32

13 96 4

73

52

y = cos(x)

ππ ππ

–

–1–

Figure 4.12

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 44

Practice

Use the unit circle and the trigonometric identities to complete the following table. Find the answers to questions

19 through 38.

–TRIGONOMETRY–

45

0° = 0 0 1 0 1 undef. undef.

30° = *19* 2

45° = 1 1

60° = 2

90° = 1 0 undef. undef. 1 0

120° = � �2 *20* �

135° = *21* *22* *23* *24* *25* *26*

150° = � � � 2

180° = π *27* �1 0 �1 *28* undef.

210° = � � � �2 *29*

225° = � � 1 1

240° = *30* *31* *32* *33* *34* *35*

270° = �1 0 undef. undef. �1 0

300° = � 2 � �

315° = � �1 *36* �1

*37* *38* �2

360° = 2π 0 1 0 1 undef. undef.

Note: The numbers appearing in bold with asterisks are questions 19 through 38.

− 32 3

3− 3

3330

11

6o = π

− 22

2

2

27

4

π

3

3

2 3

3− 31

2

3

25

3

π

3

2

π

4

3

π

− 2− 22

2

2

2

5

4

π

2 3

3

3

3

3

2

1

27

6

π

− 32 3

3

3

3

3

2

1

2

5

6

π

3

4

π

3

3− 3

1

23

2

2

3

π

π2

3

32 3

331

23

2

π3

222

2

2

2

π4

32 3

3

3

2

1

2

π6

cot1x 2csc1x 2sec1x 2tan1x 2cos1x 2sin1x 2

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 45

19. Find the value that goes in the position in the

table where you see *19*.







































Solving Simply Trigonometric Equations

The chart can be used to solve some simple equations.

ExampleFind all values of x between 0 and 2π such that

.

SolutionNote that multiples of have cosines equal to

or . Of these, the values that solve the equation

are and .

Practice

For questions 39 and 40, find the value(s) of x between

0 and 2π that satisfy the given equation.

39. .

40. cos(x) = –1.

sin( )x = − 32

x = 54πx = 3

4π

− 22

22

π4

cos( )x = − 22

–TRIGONOMETRY–

46

Calc2e_04_37-46.qxd 11/20/11 10:59 PM Page 46

LE

SS

ON

LIMITS AND CONTINUITY

The notion of a limit is the single most important underlying concept upon which calculus is built. We

can use the notion of a limit to describe the behavior of a function near a particular input, even when

the function is not defined there.

Limits can be illustrated using graphs and tables of values. For example, consider the function whose

graph is shown in Figure 5.1. We can’t talk about f(x) at x = 2 because of the unshaded circle on its graph.

But, we can talk about what happens close to 2. The values of the function at x-values close to 2 are listed in

the table.

5

47

x

)

Figure 5.1

x f(x)

1.9 5.39

1.99 5.0399

1.999 5.003999

1.9999 5.000399999

2 ???

↑ ↑2.0001 4.99959999

2.001 4.995999

2.01 4.9599

2.1 4.59

↑↑

9

–2–3 1 2 3–1

y

x

?

Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 47

The domain of f is . We can’t plug x � 2

into f. However, the hole appears to be at (2,5). How do

we know the hole has a y-value of 5? Well, the points

on the curve with x-values near x � 2 have

y-values close to y � 5. The closer we get to x � 2, the

closer the y-values of the points come to y � 5.

The mathematical shorthand for this is

= 5, which is read as “the limit as x approaches 2 of

is 5.”

The utility of limits lies in the fact that f need not

be defined at the value a in order to have a limit as x

approaches a.

We can also approach points from either the left

or from the right. For example, consider the graph of

in Figure 5.2.

Here, and .

The little minus in means that we approach

x � 1 using numbers less than (to the left) of x � 1. As

we approach x � 1 from the left-hand side, we slide up

the graph through y-values that approach 4. Similarly,

the plus in means “approach from the right.”

From the right, the height of the graph slides down to

y � 2 as x approaches 1.

In this example, does not exist because

there is no single y-value to which all of the points

near x � 1 get close. Some are close to 4, and others

are close to 2. Because there is no agreement, there is

no limit.

As another example, consider the graph of

in Figure 5.3. Here, because

sliding up to x � 3 from the left has us pass through

points with y-values near 2. Similarly, .

Because there is agreement from the left and right, we

have the general limit, . Notice that what

happens exactly at x � 3 is irrelevant. Here ,

but the resulting point at (3,5) has no bearing on the

limit as x approaches 3.

Vertical asymptotes correspond with infinite

limits. For example, consider the graph of in

Figure 5.4.

y � k1x 2

h13 2 � 5

limxS3

h1x 2 � 2

lim ( )x

h x→ +

=3

2

limxS3�

h1x 2 � 2y � h1x 2

limxS1

g1x 2

limxS1�

limx → −1

limxS1�

g1x 2 � 2limxS1�

g1x 2 � 4

y � g1x 2

f 1x 2

lim ( )x

f x→2

x � 2

–LIMITS AND CONTINUITY–

48

1

2

3

–1

–2–3 1 2 3–1

y

x

4

y = g(x)

Figure 5.2

1

2

3

4

5

6

–11 2 3–1

y

x4 5

y = h(x)

(4,1)

Figure 5.3

Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 48

Here, we write and k(x) =

. These statements simply suggest what the graph

is doing on either side of 2. The limits technically do

not exist since they are infinite.

Look back at the graphs in this lesson. Would you

agree with the following?

The difference from the limits discussed previ-

ously is that now, in each case, the function is contin-

uous at the point where we are computing the limit,

meaning that there are no holes, jumps, or asymptotes

there. Symbolically, we say f is continuous at x = a if

.

Practice

Use the graphs in Figure 5.5 to evaluate the following.

1.

2.

3.

4.

5. Is f continuous at x � �1?

6. limxS3�

f 1x 2

f 1�1 2

lim

xS �1 f 1x 2

limxS �1�

f 1x 2

limxS �1�

f 1x 2

lim ( ) lim ( ) ( )x a x a

f x f x f a→ →− +

= =

lim ( ) , lim ( ) , lim ( )x x x

f x g x h x→ →− →

= = =0 3 4

9 0 1

−∞

limx → +2

lim ( )x

k x→ −

= ∞2


49

1

2

3

–1

–2–3 1 2 3–1

y

x

4

y = k(x)

Figure 5.4

1

2

3

4

–1–2 1 2 3 4 5–1

y

x

2

3

4

5

6

–11 2 3 4 5 6–1

–2

y

x

y = f (x)

y = g(x)

1

Figure 5.5

Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 49

7.

8.

9.

10. Is f continuous at x � 1?

11.

12.

13.

14.

15.

16.

Evaluating Limits Algebraically

It is not necessary to have the graph of a function to

evaluate its limits. For instance, if f is continuous at a,

then its limit as x approaches a is simply f (a).

Technically, this works only with functions that

are polynomials like , roots like ,

rational functions (formed by dividing two poly-

nomials) like , trigonometric functions,

and transcendental functions like , and .

Because this works for any combination of these func-

tions added, subtracted, multiplied, divided, or com-

posed, it works also for every function considered in

this book.

ExampleEvaluate and .

SolutionBecause 4 can be plugged into without

there being a division by zero, the limit

� . Similarly, �

.

Practice

Evaluate the following limits.

17.

18.

19.

20.

21.

22.

Dividing by a tiny number is equivalent to mul-

tiplying by an enormous number. For example:

It is for this reason that if the denominator of a frac-

tion approaches zero while the numerator goes to a

nonzero number, the result is an infinite limit.

A classic example is (graphed in Fig-

ure 5.6).

f 1x 2 �1x

5 �1

10,000 � 5 # 10,000

1 � 50,000

limx

xe→−

−

2

3 2

lim( )a

x a→

+ +0

2 1

limxSp

6

sin1x 2

x

limx

x

xx

→

−+

+

2

324

10 3

limxS3

x � 3x2 � x

lim( )x

x x x→

+ − +1

3 210 4 5 7

31�2 22 � 1�2 2 � 7 � 3

lim ( )x

x x→−

+ −2

23 74 � 516 � 40

�9

56

limxS4

x � 5

x2 � 10x

x � 5x2 � 10x

lim ( )x

x x→−

+ −2

23 7limxS4

x � 5

x2 � 10x

exln1x 2

3x � 52x3 � x2 � 1

x4x5 � 10x3 � 7

limxS5�

g1x 2

limxS5�

g1x 2

limxS3

g1x 2

limxS3�

g1x 2

g11 2

limxS1

g1x 2

f 13 2

lim

xS3 f 1x 2

limxS3�

f 1x 2


50

Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 50

Because the denominator goes to zero while the

numerator stays one in all of these cases, there is a ver-

tical asymptote at x � 0. The function therefore

approaches either positive or negative infinity from

either side. When x is less than zero, as it always is when

, the function is also negative. Thus,

. Similarly, as , is always

positive, so . Finally, because the limit

from the two sides are different, the undirected limit

does not exist.

Example

Evaluate .

SolutionThe numerator approaches 5 while the denominator

approaches 0 as x approaches 2 from the right. There-

fore, this limit from the right is either ∞ or �∞. What

we need to determine is whether the function is posi-

tive or negative at x-values just slightly larger than 2.

We do this by looking at each factor individually.

As , the values we are plugging into each

factor are slightly larger than 2. So, (x + 3) and (x – 2)

are both positive, while (x – 4) is negative. Because the

function is made of two positive parts

and one negative part, the entire fraction will be neg-

ative. Thus, = –∞.

Because this is negative, the limit is �∞. Another

method will be covered in Lesson 13.

Example

Evaluate .

SolutionHere, the numerator approaches �10, which isn’t zero,

while the denominator approaches zero, so the limit is

either q or �q. As , the values we are plug-

ging into each factor are slightly smaller than –3. So,

(x + 1) and (x + 3) are both negative, while (2 – x) and

(x + 5) are both positive.

The combination of two negative factors and two

positive factors is positive, thus:

lim( )( )

( )( )x

x x

x x→− −

+ −+ +

= ∞3

1 2

3 5

x → − −3

limxS �3�

1x � 1 2 12 � x 2

1x � 3 2 1x � 5 2

lim( )( )x

xx x→ +

+− −2

32 4

xx x

+− −

32 4( )( )

x → +2

lim( )( )x

xx x→ +

+− −2

32 4

limxS0

1x

limx x→ +

= ∞0

1

1x

x S 0�limx x→ −

= − ∞0

1

1x

x S 0�


51

1

2

3

4

5

6

–1–2–3–4–5–6 1 2 3 4 5 6–1

–2

–3–4

–5

–6

y

x

y = 1__x

Figure 5.6

Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 51

Practice


23.

24.

25.

26.

27.

28.

When both the numerator and the denominator

go to zero, there are some common tricks for simpli-

fying the limit. The first is to factor and cancel. The

second is to rationalize. These are illustrated next.

Example

Evaluate .

SolutionHere, both the numerator and denominator go to zero,

so we aren’t guaranteed an infinite limit. First, factor

the numerator and denominator.

Because as , we can cancel .

Now we can plug in without dividing by zero.

The following example utilizes the trick of

rationalizing.

Example

Evaluate .

SolutionBecause both numerator and denominator go to zero,

a trick is necessary. First, multiply the top and bottom

by the part with the square root, but with the opposite

sign between them.

Simplify.

= lim( )x

x

x x→

−− +( )9

9

9 3

lim lim( )( )x x

x

x

x x x

x x→ →

−−

= + − −− +9 9

3

9

3 3 9

9 3

lim limx x

x

x

x

x

x

x→ →

−−

= −−

++

⋅

9 9

3

9

3

9

3

3

limx

x

x→

−−9

3

9

limxS4

1x � 2 2

1x � 5 2�

69

�23

limxS4

x2 � 2x � 8x2 � x � 20

�

lim

xS4 1x � 4 2 1x � 2 2

1x � 4 2 1x � 5 2� lim

xS4 1x � 2 2

1x � 5 2

limxS4

x2 � 2x � 8x2 � x � 20

�

x � 4x � 4

� 1x S 4x � 4

limxS4

x2 � 2x � 8x2 � x � 20

� limxS4

1x � 4 2 1x � 2 2

1x � 4 2 1x � 5 2

limxS4

x2 � 2x � 8x2 � x � 20

limxS �5

x � 21x � 5 22

limxS2

1x � 5 2 1x � 5 2

1x � 3 2 1x � 4 2

limxS3�

1x � 2 2 1x � 5 2

1x � 6 2 1x � 3 2

limx → −3

limxS4�

x � 5x � 4

limx x→ − −1

1

1


52

When computing a limit as x goes to a, if plug-ging in a results in 0

0, do NOT automatically con-clude the limit is 1. This means you must simplifythe expression somehow before plugging in a.

Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 52

Eliminate .

Plug in.

Factoring tricks can even be useful when dealing with

transcendental functions.

ExampleEvaluate

SolutionTo compute this limit, use the fact that and

factor the denominator as a difference of squares.

Then, cancel factors that are common to both the

numerator and denominator, and substitute x = 0 into

the simplified expression, as follows:

Practice


29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40. limln( )z

z z

z

e e

e→

( ) − ( ) +

−2

26 8

2

limx

x x xx→

+ −−3

3 2

2

2 159

limh

x h x

h→

+( ) −0

3 32 2

limcos( )cos( )x

xx→

+−π

11

lima

a

x a x→ + −0

limaS0

1x � a 22 � x2

a

lim(x

x

x x→

−−25

5

2

5)( + 1)

limxS3�

x � 5x � 3

limxS3

x2 � 4x � 3

x2 � 2x � 15

limxS4

x2 � 9x � 3

limxS2

x � 2x2 � 4

1x � 6 2 1x � 2 2

1x � 2 2 1x � 1 2lim

xS �2�

limx

xe e→ + +=

+=

+=

00

11

11

11 1

12

lim lim limx

x

xx

x

x x

ee

e

e→ → →+ + +−−

= −− ( )

=0

20

20

11

1

1

e ex x22

= ( )

limx

x

x

ee→ +

−−0 2

11

lim lim( )x x

x

x x→ →

−−

=+

=9 9

3

9

1

3

1

6

lim lim( )

( )( )x x

x

x

x

x x→ →

−−

= −− +9 9

3

9

9

9 3

x � 9x � 9

� 1


53

1

1 1

−−( ) +( )

e

e e

x

x x

Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 53

Calc2e_05_47-54.qxd 11/18/11 12:41 AM Page 54

LE

SS

ON

DERIVATIVES

Straight lines are convenient to deal with, but most functions have curved graphs. This does not stop

us from projecting straight lines on them! For example, at the point marked x on the graph in Figure

6.1, the function is clearly increasing. However, exactly how fast is the function increasing at that point?

Since “how fast” refers to a slope, we draw in the tangent line, the line straight through the point that heads

in the same direction as the curve (see Figure 6.2). The slope of the tangent line tells us how fast the func-

tion is increasing at the given point.

6

55

x

y = f(x)

Figure 6.1

Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 55

We can figure out the y-value of this point by

plugging x into f and getting . However, we

can’t get the slope of the tangent line when we have just

one point. To get a second point, we go ahead a little

further along the graph (see Figure 6.3). If we go ahead

by distance a, the second point will have an x-value of

and a y-value of .

Because this second point is on the curve and not

on the tangent line, we get a line that is not quite the

tangent line. Still, its slope will be close to the one we

want, so we calculate as follows:

slope �

To make things more accurate, we pick a second

point that is closer to the original point (x,f (x)) by

using a smaller a. This is depicted in Figure 6.4.

In fact, if we take the limit as a goes to zero from

the right, we will get the slope of the tangent line

exactly. The situation is completely similar if we take a

< 0 so that x + a is to the left of x. This process gives us

the derivative of and is written .

� slope of the tangent line at point

ExampleWhat is the derivative of ?

SolutionStart with the definition of the derivative.

Use .

f ¿ 1x 2 � limaS0

1x � a 22 � x2

a

f 1x 2 � x2


f 1x � a 2 � f 1x 2

a

f 1x 2 � x2

1x,f 1x 2 2

f˛˛¿ 1x 2 � limaS0

f 1x � a 2 � f 1x 2

a

f ¿ 1x 2f 1x 2

f 1x � a 2 � f 1x 2

1x � a 2 � x�

f 1x � a 2 � f 1x 2

a

f 1x � a 2x � a

1x,f ˛1x 2 2

–DERIVATIVES–

56

x

y = f(x)tangentline

Figure 6.2

x

y = f(x)tangentline

x + a

(x, f(x))

(x + a, f(x + a))

not quite thetangent line

Figure 6.3

x

y = f(x)tangentline

x + a

(x, f(x))

(x + a, f(x + a))

closer to thetangent line

Figure 6.4

Remember, f (x + a) ≠ f (x) + f (a).

Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 56

Multiply out and simplify.

Factor and simplify.

Evaluate the limit.

The derivative is . This means that

the slope at any point on the curve is exactly

twice the x-coordinate. The situation at x � �2,

x � 0, and x � 1 is shown in Figure 6.5.

ExampleWhat is the slope of the line tangent to g(x) =

at x � 9?

SolutionStart with the definition of the derivative.

Use g(x) = .

Rationalize the numerator.

g′(x) =

Multiply and simplify.

g′(x) =

Simplify.

Plug in to evaluate the limit.

The derivative of g(x) = is thus

g′(x) . This means that at x � 9, the slope of

the tangent line is g′(9) . This is illus-

trated in Figure 6.6.

ExampleFind the equation of the tangent line to

at x � 3.h1x 2 � 2x2 � 5x � 1

= = 1

2 9

1

6

= 1

2 x

x

= 1

2 x

=+ +

1

0x x

1

x a x + +g¿ 1x 2 � lim

aS0

′ = // + +→

g xa

a x a xa( ) lim

( )

0

lim( )a

x a x x a x x a x

a x a x→

+ + ⋅ + − ⋅ + −+ +0

lima

x a x

a

x a x

x a x→

+ −

+ ++ +

0

′ = + −→

g xx a x

aa( ) lim

0

x

g¿ 1x 2 � limaS0

g1x � a 2 � g1x 2

a

x

y � x2

f ¿ 1x 2 � 2x


2x � a � 2x


12x � a 2a

a


x2 � 2ax � a2 � x2

a

–DERIVATIVES–

57

1

2

3

–1

–2–3 1 2 3–1

y

x

4at (–2,4)

slope = –4

slope = 0at (0,0)

slope = 2at (1,1)

y = x2

Figure 6.5

Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 57

SolutionTo find the equation of the tangent line, we need a

point and a slope. The y-value at x � 3 is

, so the point is (3,4).

And to get the slope, we need the derivative. Start with

the definition of the derivative.

Thus, the derivative of is

. The slope at x � 3 is

. The equation of the tangent

line is therefore . This

is shown in Figure 6.7.

y � 71x � 3 2 � 4 � 7x � 17

h¿ 13 2 � 413 2 � 5 � 7

h¿ 1x 2 � 4x � 5

h1x 2 � 2x2 � 5x � 1

h¿ 1x 2 � limaS0

h1x � a 2 � h1x 2

a

h13 2 � 213 22 � 513 2 � 1 � 4

–DERIVATIVES–

58

1

2

3

1 2 3 4 5 6

y

x7 8 9 10

xy = g(x) =

slope = 1-6

at (9,3)

Figure 6.6

Use .

Multiply out and simplify.

Factor out and simplify.

Evaluate the limit.

′ = + − = −→

h x x a xa

( ) lim( )0

4 2 5 4 5

14x � 2a � 5 2a

ah¿ 1x 2 � lim

aS0

2x2 � 4ax � 2a2 � 5x � 5a � 1 � 2x2 � 5x � 1

ah¿ 1x 2 � lim

aS0

21x � a 22 � 51x � a 2 � 1 � 12x2 � 5x � 1 2

ah¿ 1x 2 � lim

aS0

h1x 2 � 2x2 � 5x � 1

Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 58

Practice

1. Find the derivative of .

2. If , then what is ?

3. Find the derivative of .

4. Find the derivative of f(x) = 3 .

5. If , then what is

6. Find the slope of at x � 2.

7. Find the slope of g(x) = at x � 16.

8. Where does the graph of

have a slope of 0?

9. Find the equation of the tangent line to

at (2,�3).


when x � 1.

Derivatives Don’t Always Exist!

Because the definition of involves a full-blown

limit in order for it to exist, the following left- and

right-hand limits must be equal:

This is not always the case, though.

ExampleLet us try to compute the derivative of A(x) = |x| at

x = 0 (shown in Figure 6.8).

lim( ) ( )

lim( ) ( )

a a

f x a f xa

f x a f xa→ →− +

+ − = + −0 0

′f x( )

k1x 2 � 5x2 � 2x

h1x 2 � 1 � x2

g1x 2 � x2 � 4x � 1

3 x

f 1x 2 � 3x2 � x

k¿ 1x 2?k1x 2 � x3

x

g1x 2 � 10

h¿ 1x 2h1x 2 � x2 � 5

f 1x 2 � 8x � 2

59

–2 1 2–1

y

x

1

2

3

4

5

6

7

8

9

10

3 4–1

–2

(3,4)

h(x) = 2x2 – 5x + 1

y = 7x – 17

Figure 6.7

The absolute value of x, denoted x , tells you how far x is from zero. For instance, 5 = 5 (since 5 is five unitsfrom 0 on the right-hand side) and |–4| = 4 (since –4 is four units from 0, just on the left-hand side). Symboli-cally,

x =x, whenever x ≥ 0

–x, whenever x < 0

ABSOLUTE VALUE

Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 59

For a > 0, we have

So, .

Now, let us take a < 0. Doing so, we get:

So, .

But, the left- and right-hand limits aren’t equal. So,

the full-blown limit does not

exist. Thus, m′(0) doesn’t exist. Geometrically, the

sharp corner in the graph of y = m(x) at x = 0 is the rea-

son the derivative doesn’t exist there.

lim( ) ( )

a

m a ma→

+ −0

0 0

lim( ) ( )

lim ( )a a

m a ma→ →− −

+ − = − = −0 0

0 01 1

m a ma

aa

aa

( ) ( )0 0 01

+ − =−

= − = −

lim( ) ( )

lima a

m a ma→ →+ +

+ − = =0 0

0 01 1

m a ma

aa

aa

( ) ( )0 0 01

+ − =−

= =

–a ac

y

x

slope–1

m(x) = x

slope1

–DERIVATIVES–

60

Calc2e_06_55-60.qxd 11/18/11 12:43 AM Page 60

LE

SS

ON

BASIC RULES OF DIFFERENTIATION

Using the limit definition to find derivatives can be very tedious. Luckily, there are many shortcuts

available. For example, if function f is a constant, like or , then . This

can be proven for all constants c at the same time in the following manner.

If:

then:

All of the general rules in this chapter can be proven in such a manner, using the limit definition of the deriv-

ative, though we shall not actually do so. The first rule is the Constant Rule, which says that if where

c is a constant, then .

Before we go any further, a word needs to be said about notation. The concept of the derivative was dis-

covered by both Isaac Newton and Gottfried Leibniz. Newton would put a dot over a quantity to represent

its derivative, much like we have used the prime notation to represent the derivative of . Leibniz

would write the derivative of y (where x is the variable) as . Newton’s notation is certainly moredy

dx

f 1x 2f ¿ 1x 2

f ¿ 1x 2 � 0

f 1x 2 � c

f ¿ 1x 2 � lim

aS0 f 1x � a 2 � f 1x 2

a� lim

aS0 c � c

a� lim

aS0 0a

� 0

f 1x 2 � c

f ¿ 1x 2 � 0f 1x 2 � 18f 1x 2 � 5

7

61

Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 61

convenient, but Leibniz’s enables us to represent

“take the derivative of something” as (something).

Thus, if , then .

Using Leibniz’s notation, the Constant Rule where c is

a constant is expressed as .

The next rule is the Power Rule, which is stated:

. This rule says “multiply the expo-

nent in front and then subtract one from it.”

ExampleDifferentiate .

Solution


Solution


SolutionTo use the Power Rule, we need expressed as x

raised to a power, or:

Notice how much easier it is to use the Power

Rule to compute the derivative than it was using the

limit definition of the derivative in Lesson 6.

′ = = = =− − ⋅g x x x

x x( )

1

2

1

2

1

2

1

2

12

12

1 1

g1x 2 � x12

g1x 2

g x x( ) =

dy

dxx x = =−8 88 1 7

y � x8

f ¿ 1x 2 � 2x2�1 � 2x1 � 2x

f 1x 2 � x2

ddx1xn 2 � n # xn�1

ddx1c 2 � 0

ddx1f 1x 2 2 � f ¿ 1x 2

dy

dx �y � f 1x 2

ddx

62

CONSTANT RULE

If f (x) = c where c is a constant, then f ′(x) = 0.

And, using Leibniz’s notation, if c is a constant, then .ddx

c( ) = 0

POWER RULE

ddx

x n xn n( ) = ⋅ −1

Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 62

Example

Differentiate .

SolutionFirst, rewrite y as so that it becomes x raised to a

power. Then,

Notice that means “take the derivative with

respect to variable t.” While x is often used as the vari-

able, so the derivative of is ,

sometimes it is convenient to use other variables. If

, then is the derivative of f with

respect to u, for example.


Solution


SolutionRewrite using the exponent rules.

Now, use the Power Rule to differentiate.

Practice

Differentiate each of the following.

1.

2. y =

3.

4.

5. y = t12

6.

7.

8.

9.

10.

11.

12.

13. f xx

( ) 1=

y �1x

y u =

k1x 2 � 24 x

g1x 2 � x�45

f 1t 2 � �11

f 1x 2 � x100

y � x 75

h1x 2 � 8

g1u 2 � u�5

x21

f 1x 2 � x5

dydt

t= − −52

72

yt

ttt

t t t= = = ⋅ =− −3 3

3

12 1

252

yt

t=

3

ddt123 t 2 �

ddt1t

13 2 �

13

t 13 �1 �

13

t�23 �

1

3t 23

y t= 3

dy

du� f ¿ 1u 2y � f 1u 2

dy

dx� f ¿ 1x 2y � f 1x 2

ddt

�ddx1x�2 2 � �2x�2�1 � �2x�3 �

�2x3

dy

dx�

ddxa

1x2b

x�2

y �1x2

63

THE CONSTANT COEFFICIENT RULE

If a function has a constant multiplied in front, leave it while you take the derivative of the rest.Using Leibniz’s notation, , where c is a constant.d

dxcf x c

ddx

f x( ( )) ( ( )) =

Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 63

–BASIC RULES OF DIFFERENTIATION–

14.

15.

16.

The Constant Coefficient Rule

The Constant Coefficient Rule is stated as follows: If a

function has a constant multiplied in front, leave it

while you take the derivative of the rest. This means

that because , the derivative of

would be 5 • (8x7) = 40x7. Just imagine that the constant

steps aside and waits while you differentiate the rest.

ExamplesDifferentiate the following.

Solutions

f ′(x) = 11 ⋅ (4x3) = 44x3

(1) = 12

In that last example problem, don’t forget that p

is a constant, and thus should be treated just as

you would or .

The Additive Rule

Next, we will examine the Additive Rule, which says

that if parts of a function are added together, dif-

ferentiate the parts separately and add the results. We

know that and . The

Additive Rule then says that if , then

.


Solution′ = + = +f x

ddx

xddx

x x x( ) ( ) ( )4 30 20 605 2 4

f 1x 2 � 4x5 � 30x2

20x � 12

ddx110x2 2 �

ddx112x 2 �

dy

dx�

ddx110x2 � 12x 2 �

y � 10x2 � 12x

ddx112x 2 � 12

ddx110x2 2 � 20x

712r20r

2p r

′ = ⋅ =A r r r( ) ( )π π2 2

′ = ⋅

= =− −k u u u

u( )

154

13

54

5

4

23

23

23

dy

dx� 12

′ = ⋅ −( ) = − = −− −h t t tt

( ) 4 6 24247 7

7

′ = ⋅

= =− −g x x x

x( ) 3

12

32

3

2

12

12

dy

dxx x= ⋅ =1 20 (2 0)

A1r 2 � p r2

k1u 2 �1523 u

4�

154

u13

y � 12x

h tt

t( ) = = −44

66

g x x x( ) = =3 312

y x= 10 2

f x x( ) = 11 4

5x8ddx1x8 2 � 8x7

yt

t=

3

4

h ttt

( ) = −

5

2

g xx x

( ) = 1

64

Don’t make the mistake that

.ddx

c f xddx

cddx

f x( ( )) ( ) ( ( ))

⋅ = ⋅ =

= 0

0123

Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 64


SolutionThis can be rewritten as a sum:

g(x) = x3 + (–4)x2

thus:

The previous example shows that the Additive

Rule applies to cases of subtraction as well.


Solutions

Practice

Differentiate the following.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29. y � 3 �2x

�1x2

h1u 2 � u5 � 4u4 � 7u3 � 2u2 � 8u � 2

g1x 2 � 3x15 � 5x3

F1x 2 � 6x100 � 10x50 � 4x25 � 2x10 � 9

s t t et t( ) ln( )= + + +π 3 2 3 3

y x x x= − −− −4 32

f 1x 2 � 8x3 � 3x2

4 3 703 2t t− +

k1x 2 � 1 � x2

g1t 2 �12t4

5

V1r 2 � 43p r3

f 1x 2 ��3x10

y x= 6 7

′ = − + = −− −k t t t

t t( )

125

2 012

5

215 2

5 2

h¿ 1x 2 � 40x4 � 40x3 � 9x2 � 14x � 5

dy

dx x x

1

2 +

1

2= =0

k t tt

e t t e( ) = + + = + +−32

3 245

45 1

h1x 2 � 8x5 � 10x4 � 3x3 � 7x2 � 5x � 4

y x x + + = =4 412

′ = ( ) + −( ) = + − ⋅

= −

g xddx

xddx

x x x

x x

( ) ( )

.

3 2 2

2

4 3 4 2

3 8

g1x 2 � x3 � 4x2

65

THE ADDITIVE RULE

If parts of a function are added together, differentiate the parts separately. Then, add the results. Using Leibniz’s notation, .

ddx

f x g xddx

f xddx

g x( ( ) ( )) ( ( )) ( ( )) + = +

Remember, e is a number (approximately equal to2.71828…).

Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 65

30.

31.

32.

The derivative of the derivative is called the sec-

ond derivative. The derivative of that is the third deriv-

ative, and so on. Using notation, if y = f(x), then the

derivative is , the second derivative

is , the third derivative is ,

and the tenth derivative, for example, is .

We put the 10 in parentheses because counting the ten

primes in is ridiculous.

ExampleFind the first three derivatives of y = .

Solution

When working on multiple derivatives like this,

it makes sense to leave the exponents negative and

fractional until all derivatives have been computed.

ExampleFind all the derivatives of .

Solution

All of the subsequent derivatives will also be zero, so

we can write

for .

Practice

33. Find the first four derivatives of .

34. Find the second derivative of s(t) = – 16t 2 +

at + b, where a and b are constants.

35. Find the third derivative of

.

36. Find the first three derivatives of .y � 623 t

y x x x= − +− − −3 23 2 1

f 1x 2 �1x

n � 4f ˛

1n2 1x 2 � 0

f 1x 2 � 0

f ‡ 1x 2 � 6

f – 1x 2 � 6x � 8

f ¿ 1x 2 � 3x2 � 8x � 5

f 1x 2 � x3 � 4x2 � 5x � 7

f 1x 2 � x3 � 4x2 � 5x � 7

d 3y

dx3 �38

x�52

d 2y

dx2 � �14

x�32

dy

dx�

12

x�12

y � x 12

x

f –1x 2

d10y

dx10 � f 1102 1x 2

d3y

dx3 � f ˛‡ 1x 2d2y

dx2 � f – 1x 2

dy

dx� f ¿ 1x 2

f xx x x

( ) = −−3

25

4

y x x 4 + = =9 93

y � u2 � u�2

–BASIC RULES OF DIFFERENTIATION–

66

Calc2e_07_61-66.qxd 11/18/11 12:44 AM Page 66

LE

SS

ON

RATES OF CHANGE

It is useful to contemplate slopes in practical situations. For example, suppose the following graph in Fig-

ure 8.1 is for , a function that gives the price y for various amounts x of cheese. Because the

straight line goes through the points (1 lb.,$2) and (2 lbs.,$4), the slope � = = $2 per

pound.

$21 lb.

$ $4 22 1

−− lbs. lb.

y � f 1x 2

8

67

1

2

3

4

5

1 2 3

y

y = f(x)

Amount (in pounds)

(in dollars)

Costs of Cheese

x

6

Price

Figure 8.1

Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 67

The slope is therefore the rate at which the

price of cheese changes per pound. Because slope �

, a slope will always be a rate measured in

y-units per x-unit.

For example, suppose a passenger on a bus writes

down the exact time she passes each highway mile

marker. She then sketches the graph shown in Figure

8.2 of the bus’s position on the highway over time. The

slope at any point on this graph will be measured in y-

units per t-unit, or miles per hour. The steepness of the

slope represents the speed of the bus.

Practice

For each of the following four graphs, describe the rate

that a slope of the curve represents.

1.

2.

–RATES OF CHANGE–

68

y

y = s(t)

50

100

150

200

250

300

Posi

tion

Giv

en b

y M

arke

r (i

n m

iles)

Time on Bus (in hours)

Mile Markers

t

1 2 3

Figure 8.2

y

Time Worked (in hours)

t

1 2

Money Earned

Pay

(in

dol

lars

)

10

20

30

40

50

60

3 4 5 6 7 8 9 10

y

1 2

Gasoline Use

10

20

30

40

50

3 4

60

x

5

Dis

tan

ce D

rive

n (

in m

iles)

Gasoline Used (in gallons)

y-change

x-change

Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 68

3.

4.

Because the derivative of a function gives the

slope of its tangent lines and the tangent line indicates

how fast the function is increasing or decreasing, these

practice problems show that the derivative of a func-

tion gives its rate of change.

An excellent example comes from position func-

tions. A position function states the position of an

object along a straight line at any given time. The

derivative states the rate at which that object’s

position is changing—that is, the velocity of the func-

tion. Thus, s ′(t) = v(t). The second derivative

tells how fast the velocity is changing, or

the acceleration. Thus, s ′′(t) = v ′(t) = a(t) where s(t) is

the position function, v(t) is the velocity function, and

a(t) is the acceleration function.

ExampleSuppose an object rolls along beside a tape measure so

that after t seconds, it is next to the inch marked

. Where is the object after 1 sec-

ond? After 3 seconds? What is the velocity function?

How fast is the object moving after 2 seconds? What is

the acceleration function?

SolutionThe position function tells us

where the object is. After 1 second, the object is next to

the s(1) = 17-inch mark on the tape measure. After 3

seconds, the object is at the -inch mark.

The velocity function is .

Thus, after 2 seconds, the object is moving at the rate

of inches per second. Realize that this

velocity of 24 inches per second is an instantaneous

velocity, the speed just at a single moment. If a car’s

speedometer reads 60 miles per hour, this does not

mean that it will drive for 60 miles or even for a full

hour. The car might speed up, slow down, or stop.

However, at that instant, the car is traveling at a rate

that, if unchanged, will take it 60 miles in one hour. A

derivative is always an instantaneous rate, telling you

v12 2 � 24

v1t 2 � s¿ 1t 2 � 8t � 8

s13 2 � 65

s1t 2 � 4t2 � 8t � 5

s1t 2 � 4t2 � 8t � 5

s– 1t 2 � v¿ 1t 2

s¿ 1t 2

s1t 2


69

yGrowth of a Baby

10

20

30

x

Age (in months)

40

Wei

ght

(in

pou

nds

)

6 12 18 24 30 36

Time (in hours)

Size of Snowball(on a 40°F day)

5

y

t

Dia

met

er(i

n in

ches

)

Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 69

the slope at a particular point, but not making any

promises about what will happen next.

The acceleration function is

. Because this is a constant, it tells us that

the object increases in speed by 8 inches per second

every second.

The most popular example of constant accelera-

tion is gravity, which accelerates objects downward

by every second. Because of this, an object

dropped with an initial velocity of b feet per second

from a height of h feet above the ground will have (after

t seconds) a height of feet.

The starting time is , at which point the

object is feet off the ground, the correct

initial height. The velocity function is

. At the starting time t � 0, the velocity is

, the desired initial velocity. The function

means that 32 feet per second are

subtracted from the initial velocity b every second. The

acceleration function is .

This is the desired constant acceleration.

Note: A negative velocity means the object is moving

backward (or in the direction of decreasing y-value).

The speed is the absolute value of the velocity.

ExampleSuppose a brick is thrown straight upward with an ini-

tial speed of from a 150-foot rooftop. What are

its position, velocity, and acceleration functions?

Solution

Because the initial velocity is and the ini-

tial height is h � 150 feet, the position function is s(t)

= –16t 2 + 10t + 150. The velocity function is b(t) =

s′(t) = –32t + 10. The acceleration is a(t) = –32, a con-

stant 32 feet per second downward each second. The

negative sign indicates that gravity is acting to decrease

the height of the brick, pulling it downward.

ExampleSuppose a rock is dropped from a 144-foot tall bridge.

When will the rock hit the water? How fast will it be

going then?

SolutionBecause the rock is dropped, the initial velocity is

b � 0. The initial height is h � 144. Thus,

gives the height function. The

rock will hit the water (have a height of zero) when:

And because �3 seconds doesn’t make any sense, the

rock will hit the water after 3 seconds.

The velocity function is ;

therefore, the rock will have a velocity of

after 3 seconds. This means that it will be traveling at

a rate of 96 feet per second downward when it hits

the water.

Example

If gives the value, in

thousands of dollars, of a start-up company after t

days, then how fast is its value changing after 30 days?

After 500 days?

p1t 2 �t2

10� 80t � 50,000

v13 2 � �96

v1t 2 � s¿ 1t 2 � �32t

t � ;3

144 � 16t2

�16t2 � 144 � 0

s1t 2 � �16t2 � 144

b � 10ft

sec

10ft

sec

s– 1t 2 � �32a1t 2 � v ¿ 1t 2 �

v1t 2 � �32t � b

v10 2 � b

� �32t � b

� s¿ 1t 2v1t 2

s10 2 � h

t � 0

s1t 2 � �16t2 � bt � h

32ft

sec

s– 1t 2 � 8

v¿ 1t 2 �a1t 2 �


70

Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 70

Solution

The derivative gives the rate of

change in value, measured in thousands of dollars per

day. After 30 days, , so the company

will be losing value at a rate of $74,000 per day. After

500 days, , so the company will be gain-

ing value at the instantaneous rate of $20,000 a day.

Practice

5. The height of a tree after t years is h(t) = 30 –

feet when . How fast is the tree growing

after 5 years?

6. The level of a river t days after a heavy

rainstorm is feet. How

fast is the river’s level changing after 7 days?

7. When a company makes and sells x cars, its

profit is dollars.

How fast is its profit changing when the com-

pany makes 50 cars?

8. When a container is made x inches wide, it costs

dollars to make. How is the

cost changing when x � 3 inches?

9. An electron in a particle accelerator is

meters from the start

after t seconds. Where is it after 3 seconds? How

fast is it moving then? How fast is it accelerating

then?

10. A brick is dropped from 64 feet above the

ground. What is its position function? What is its

velocity function? What is its acceleration? When

will it hit the ground? How fast will it be travel-

ing then?

11. A bullet is fired directly upward at 800 feet per

second from the ground. How high is it when it

stops rising and starts to fall?

12. A rock is thrown 10 feet per second down a

1,000-foot cliff. How far has it gone down in the

first 4 seconds? How fast is it traveling then?

Derivatives of Sine and Cosine

Examining rates and slopes at various points can help

us determine the derivative of . Look at the

slopes at various points on its graph in Figure 8.2. It

appears that the derivative function of must

oscillate between �1 and 1, and must go through the

following points (see Figure 8.3). The function

is exactly such an oscillating function (see Figure 8.4).

This suggests that .

A similar study of the slopes of would

show that . The slopes of the

cosine function are not the values of the sine function,

but rather their exact negatives. These two results

could be obtained using the limit definition of the

derivative, but involve the use of trigonometric

identities and certain limit formulas not covered in

this book.

ExamplesDifferentiate the following

y � 2 � cos1t 2

f 1x 2 � 5sin1x 2 � 4x2

ddx1cos1x 2 2 � �sin1x 2

cos1x 2

ddx1sin1x 2 2 � cos1x 2

cos1x 2

sin1x 2

sin1x 2

s1t 2 � t3 � 2t2 � 10t

C1x 2 � 0.8x2 �24x

P xx

x x( ) , = − +3

2

1060 9 000

L1t 2 � �t2 � 8t � 26

t � 125t

p¿ 1500 2 � 20

p¿ 130 2 � �74

p¿ 1t 2 �t5

� 80


71

, not cos(c), when c is a

constant. Similarly, .ddx

ccos( ) = 0

ddx

csin( ) = 0

Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 71


72

1

–1

5π–

at x =slope = 0 5π–2

at x =slope = 1

2π

slope = 03π–2

at x =

2π3π–2slope = –1

at x = π

π π2–slope = 1

at x = 0–π

2–

slope = 0at x = – π

2–

y = sin(x) slope = 0at x =π2–

2

Figure 8.2

,0)(3π–2

�1

π2

π– 3π–22π–π

2–5π–2

(–π2–,0) ( ,0)

( ,�1)

,0)(π2–

π

2π( ,1)

5π–2

y =ddx

(sin(x)) = slopes of sin(x)1 (0,1)

Figure 8.3

π2–

�1

1

π2

π– 3π–

22π 5π–

2

y = cos(x)

Figure 8.4

Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 72

Solutions

Practice

For questions 13 through 17, compute the derivative.

13.

14.

15. g(x) =

16.

17.


at x = .

Derivatives of the Exponential and Natural

Logarithm Functions

The reason why the nicest exponential function is

where e � 2.71828 . . . is because this makes for the

following very nice derivative:

It is only with this exact base that the derivative of the

exponential function is itself (see Figure 8.5).

The derivative of its inverse function is as

follows:

A proof of this formula is given in Lesson 11.

ddx1 ln1x 2 2 �

1x

ln1x 2

ddx1ex 2 � ex

ex

π2

f 1x 2 � sin1x 2 � cos1x 2

h1x 2 � cos1x 2 � cos15 2

r( ) sin( ) cos( )θ θ θ = +1

2

1

2

352

3x

xx− + − πcos( )

f 1t 2 � 3sin1t 2 �2t

y � 4x5 � 10cos1x 2 � 3

g¿ 1x 2 � cos1x 2 � sin1x 2

dy

dt� �sin1t 2

f ¿ 1x 2 � 5cos1x 2 � 8x

g x x x( ) sin( ) cos( ) sin= − +

π6


73

1

2

3

4

5

6

–1

–2 1 2 3–1

y

x

7

8

9

y = ex

(1,e)

(2,e2)at

slope = e2

atslope = e

atslope = 1

at

slope =

(0,1)–1, 1—e1—e

Figure 8.5

Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 73


Solutions

Practice


19.

20.

21.

22.

23.

24. Find the second derivative of

.

25. Find the 100th derivative of .

26. What is the slope of the tangent line to

at x � 10?f 1x 2 � ln1x 2

g1x 2 � 3ex

f 1x 2 � ex � ln1x 2

k x x e x( ) ln( )= + +3 552 π

h x x x e( ) ln( )= − + −8 3

y � cos1x 2 � 10ex � 8x

g1t 2 � 12ln1t 2 � t2 � 4

f 1x 2 � 1 � x � x2 � x3 � ex

dydx

= 0

dydu u

e eu= − +8

g ¿ 1t 2 � 3et �2t

dy

dx� 10ex

f ¿ 1x 2 � 4ex

y u e eu

y e

u= − +

= +

8

53

ln( )

ln( )

g1t 2 � 3et � 2ln1t 2

y � 10ex � 10

f 1x 2 � 4ex


74

ex (e raised to the power x) and ex (the numbere times x) are very different functions. Note that

, whereas .ddx

ex e( ) =ddx

e ex x( ) =

and whenever c is

a positive constant.

ddx

c(ln( )) = 0ddx

ec( ) = 0

Calc2e_08_67-74.qxd 11/18/11 12:45 AM Page 74

The Product Rule

When a function consists of parts that are added together, it is easy to take its derivative: Simply take the deriv-

ative of each part and add them together. We are inclined to try the same trick when the parts are multiplied

together, but it does not work.

For example, we know that and . The derivative of their product is

. This shows that the derivative of a product is not the product of the derivatives:

Instead, we take the derivative of each part, multiply by the other part left alone, and add these results together:

This time, we did get the correct answer.

ddx1x2 # x3 2 �

ddx1x2 2 # x3 �

ddx1x3 2 # x2 � 12x 2 # x3 � 13x2 2 # x2 � 5x4

5x4 �ddx1x2 # x3 2 �

ddx1x2 2 # d

dx1x3 2 � 12x 2 # 13x2 2 � 6x3

ddx1x2 # x3 2 �

ddx1x5 2 � 5x4

ddx1x3 2 � 3x2d

dx1x2 2 � 2x

LE

SS

ON

THE PRODUCT AND QUOTIENT RULES9

75

Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 75


SolutionHere, the “first part” is and the “second part” is

. Thus, by using the Product Rule,

�

. This could be simplified as

.


Solution

Thus, the derivative is:

Example

Differentiate .

Solution

Using the product rule with can be a little bit

confusing because there is no difference between the

derivative of and “left alone.” Still, if you write

everything out, the correct answer should fall into

place, even if it looks weird.

ExampleDifferentiate y = .

Solution


SolutionWe’ll use the Product Rule with as the first part and

as the second part. However, in takingsin1x 2cos1x 2

x5

y � x5sin1x 2cos1x 2

dydt

ddt

t tddt

t t

t tt

t

tt

t

t

t

=

⋅ + ( ) ⋅

= ⋅ +

⋅

= ⋅ +

= +

−

13

13

23

13

23

23

23

13

1

1

3

1

13

1

ln( ) ln( )

ln( )

ln( )

ln( )

t t3 ⋅ ln( )

exex

ex

� 35x6 # ex � ex # 5x7 � 5x6ex17 � x 2

ddx15x7 2 # ex �

ddx1ex 2 # 5x7g¿ 1x 2 �

g1x 2 � 5x7 # ex

′ = −f xx

xx x( )

cos( )sin( )ln( )

�1x

# cos1x 2 � sin1x 2 # ln1x 2

ddx1 ln1x 2 2 # cos1x 2 �

ddx1cos1x 2 2 # ln1x 2f ¿ 1x 2 �

f 1x 2 � ln1x 2 # cos1x 2

dy

dx� x213sin1x 2 � xcos1x 2 2

3x2sin1x 2 � cos1x 2 # x3

ddx1x3sin1x 2 2 �

ddx1x3 2 # sin1x 2 �

ddx1sin1x 2 2 # x3

sin1x 2

x3

y � x3sin1x 2

76

THE PRODUCT RULE

The Product Rule can be stated “the derivative of the first times the second, plus the derivative of the secondtimes the first.” It can be proven directly from the limit definition of the derivative, but only with a few tricks anda lot of algebra. Using Leibniz’s notation, the Product Rule is stated as follows:

ddx

f x g x f x g x g x f x( ( ) ( ) ( ) ( ) ( ) ( )) =⋅ ′ ⋅ + ′ ⋅

Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 76

the derivative of , we’ll have to use the

Product Rule a second time. This can get messy, but it

will be fine if everything is written down carefully.

Practice


1.

2.

3. y =

4.

5.

6.

7.

8. h(t) =

9.

10.

11. . (Hint: .)

12.


at (0,2)?


at x = π.

The Quotient Rule

The Quotient Rule for functions where the parts are

divided is slightly more complicated than the Product

Rule. The Quotient Rule can be stated:

Just as with the Product Rule, each part is differ-

entiated and multiplied by the other part. Here, how-

ever, they are subtracted, so it matters which one is

differentiated first. It is important to start with the

derivative of the top.

ddxa

f 1x 2

g1x 2b �

f˛ ¿ 1x 2g1x 2 � g¿ 1x 2f 1x 2

1g1x 2 22

y � xsin1x 2

f 1x 2 � x2ex � x � 2

g1x 2 � 3x4ln1x 2cos1x 2

y x e xx= ⋅( )sin( )y � xexsin1x 2

f 1x 2 � sin21x 2 � sin1x 2 # sin1x 2

y � 5x3 � xln1x 2

t t t3 4+( ) −( )sin( ) cos( )

y � 8ln1x 2sin1x 2 � cos1x 2

k x x xx

x( ) cos( )

sin( )= −4

h u u u eu( ) = +( )2 3

g1x 2 � 3x2ln1x 2 � 5x4 � 10

π sin( )cos( )x x

y � 8t3et

f 1x 2 � x2cos1x 2

= + − ⋅4 5 2 2 5x x x x x xsin( )cos( ) (cos ( ) sin ( ))

Bcos1x 2 # cos1x 2 � sin1x 2 # sin1x 2R # x5

= 4 5x x xsin( )cos( )

ddx1cos1x 2 2 # sin1x 2 R # x5

B ddx1sin1x 2 2 # cos1x 2 �= 4 5x x xsin( )cos( )

dy

dx�

ddx1x5 2 # sin1x 2cos1x 2 �

ddx1sin1x 2cos1x 2 2 # x5

sin1x 2cos1x 2

77

THE QUOTIENT RULE

ddx

f xg x

f x g x g x f x

g

( )( )

( ) ( ) ( ) ( )

= ′ − ′

( (x))2

Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 77

Example

Differentiate .

Solution

Here, the top part is and the bottom

part is . Therefore, by the Quotient Rule:

=

=

Example

Differentiate .

Solution

Example

Differentiate .

SolutionHere, the Product Rule is necessary to differentiate

the top.

Example

Differentiate .

Solution

= −12

ln( )t

t

=⋅ − ⋅1

1

2t

t t

t

ln( )

dy

dt�

ddt1 ln1t 2 2 # t � d

dt 1t 2 # ln1t 2

t2

y �ln1t 2

t

=⋅ + ⋅[ ]⋅ −2 2

2

x x x x x x x

x

sin( ) cos( ) ln( ) sin( )

(ln( ))

dy

dx�

ddx 1x

2sin1x 2 2 # ln1x 2 � ddx 1 ln1x 2 2 # x2sin1x 2

1 ln1x 2 22

y �x2sin1x 2

ln1x 2

= − −−

= −−

30 3 2010 1

10 310 1

4 2 4

2 2

4 2

2 2

x x xx

x xx( ) ( )

= ⋅ − ⋅−

( ) ( )( )

( )

3 10 1 20

10 1

2 2 3

2

x x x x

x

f ¿ 1x 2 �ddx 1x

3 2 # 110x2 � 1 2 � ddx 110x2 � 1 2 # x3

110x2 � 1 22

f 1x 2 �x3

10x2 � 1

15x4 � 6x 2 # cos1x 2 � sin1x 2 # 1x5 � 3x2 � 1 2

cos21x 2

15x4 � 6x 2 # cos1x 2 � 1�sin1x 2 2 # 1x5 � 3x2 � 1 2

cos21x 2

dydx

x x x x x x

x

ddx

ddx

=

− + ⋅ − ⋅ − +( ) cos( ) (cos( )) ( )

(cos( ))

5 2 5 2

2

3 1 3 1

cos1x 2

x5 � 3x2 � 1

y �x5 � 3x2 � 1

cos1x 2

–THE PRODUCT AND QUOTIENT RULES–

78

Just as with the Product Rule,

.ddx

f xg x

f x

g x

ddxddx

( )( )

( ( ))

( ( ))

≠

=⋅ + ⋅

⋅ − ⋅d

dxd

dx xx x x x x x x

x

( ) sin( ) (sin( )) ln( ) sin( )

(ln( ))

2 2 1 2

2

Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 78

Practice


15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27. Find the second derivative of y = xex + ex.


f(x) = x2 ln(x) at x = e?

Derivatives of Trigonometric Functions

We can find the derivatives of the rest of the trigono-

metric functions using the Quotient Rule.


Solution

Use .

Differentiate using the Quotient Rule.

Simplify.

Use .

Use .

Thus:ddx1tan1x 2 2 � sec21x 2

dy

dx� sec21x 2

sec1x 2 �1

cos1x 2

dy

dx�

1cos21x 2

sin21x 2 � cos21x 2 � 1

dy

dx�

cos21x 2 � sin21x 2

cos21x 2

dy

dx�

cos1x 2 # cos1x 2 � 1�sin1x 2 2 # sin1x 2

cos21x 2

dy

dx�

ddx1tan1x 2 2 �

ddxa

sin1x 2

cos1x 2b

tan1x 2 �sin1x 2

cos1x 2

y � tan1x 2

f 1x 2 �x2ex

cos1x 2

y �xln1x 2

ex

h tt t

t( )

ln( )

sin ( )= +

2

yx x

=+( ) −( )

1sin( ) cos( )π π

g uu

u u( )

sin( )=−3 3

yx x

x x= −

+1

1

g t tt

( )sin( )

=3

π

y �4et � t

t3 � 2t � 1

y �x5

ln1x 2

f 1x 2 �x � ln1x 2

ex � 1

f 1x 2 �x2 � 1x2 � 1

h1x 2 �x3 � 10x � 73x2 � 5x � 2


79

Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 79


Solution

Use .

Differentiate using the Quotient Rule.

Simplify.

Use and .

Thus:

Practice


29.

30.

31.

32. g(x) = ex sec(x)

33. h(t) = et ln(t)tan(t)

34. j xx x

x( )

sec( )= +3 4

f 1x 2 � xtan1x 2

y � cot1x 2

y � csc1x 2

ddx1sec1x 2 2 � sec1x 2tan1x 2

dy

dx� sec1x 2tan1x 2

tan1x 2 �sin1x 2

cos1x 2sec1x 2 �

1cos1x 2

dy

dx�

sin1x 2

cos21x 2�

1cos1x 2

# sin1x 2

cos1x 2

dy

dx�

0 # cos1x 2 � 1�sin1x 2 2 # 1

cos21x 2

dy

dx�

ddx1sec1x 2 2 �

ddxa

1cos1x 2

b

sec1x 2 �1

cos1x 2

y � sec1x 2


80

et ⋅ ln(t) and eln(t) are NOT the same function. Thefirst one is a product, whereas the second is acomposition.

Calc2e_09_75-80.qxd 11/18/11 12:47 AM Page 80

LE

SS

ON

CHAIN RULE

We have learned how to compute derivatives of functions that are added, subtracted, multiplied,

and divided. Next, we will learn how to compute the derivative of a composition of functions

For example, it would be difficult to multiply out just to take the

derivative. Instead, notice that looks like put inside . Therefore, in

terms of composition, f(x) = (h ° g)(x) = h(g(x)).

The trick to differentiating composed functions is to take the derivative of the outermost layer first, while

leaving the inner part alone, and then multiplying that by the derivative of the inside.

Using Leibniz’s notation, the Chain Rule can be stated as follows:

If this is confusing, think of the Chain Rule in the following way:

ddx

h1something 2 � h¿ 1something 2 # ddx1something 2

ddx1h1g1x 2 2 2 � h¿ 1g1x 2 2 # g¿ 1x 2

h1x 2 � x5g1x 2 � x3 � 10x � 4f 1x 2

f 1x 2 � 1x3 � 10x � 4 25

10

81

Calc2e_10_81-84.qxd 11/18/11 12:48 AM Page 81

The usual key to figuring out what is inside and

what is outside is to watch the parentheses. Imagine

that the parentheses form the layers of an onion, and

that you must peel (differentiate) the outermost layers

one at a time before reaching the inside.

Example

Differentiate .

Solution

Here, where the something �

. Because , the Chain

Rule gives:

= 5(x3 + 10x + 4)4 ⋅ (x3 + 10x + 4)

= 5(x3 + 10x + 4)4 ⋅ (3x2 + 10)


SolutionHere, the function is essentially sin(something) where

the “something” � . The deriva-

tive of sine is cosine, so the Chain Rule gives:


SolutionThis is tricky because of the lack of parentheses. It

might look like the “outside” function is cos(some-

thing), but it is actually .

Thus, this function is really . So, the

Chain Rule gives:

= 3(cos)(x))2 ⋅ (cos(x))

= 3(cos)(x))2 ⋅ (–sin(x))

= –3(cos)2 ⋅ (x)sin(x)

ExampleDifferentiate .y � cos1x3 2

d

dx

dy

dx� 31something 22 # d

dx1something 2

1something 23y � cos31x 2 � 1cos1x 2 23

y � cos31x 2

132x3 � 6x � 2 2

g¿ 1x 2 � cos18x4 � 3x2 � 2x � 1 2 #

ddx18x4 � 3x2 � 2x � 1 2

g¿ 1x 2 � cos18x4 � 3x2 � 2x � 1 2 #

g¿ 1x 2 � cos1something 2 # ddx1something 2

8x4 � 3x2 � 2x � 1

g1x 2 � sin18x4 � 3x2 � 2x � 1 2

d

dx

f ¿ 1x 2 � 51something 24 # ddx1something 2

ddx1x5 2 � 5x4x3 � 10x � 4

f 1x 2 � 1something 25

f 1x 2 � 1x3 � 10x � 4 25

82

THE CHAIN RULE

or ddx

h hddx

( (something)) (something) something= ′ ⋅ ( )ddx

h g x h g x g x( ( ( ))) ( ( )) = ′ ⋅ ′( )

Make certain to not mix the derivatives of the lay-ers to get and mistakenly say f ′(x) = 5(3x2 + 10)4.

Calc2e_10_81-84.qxd 11/18/11 12:48 AM Page 82

SolutionIn this example, our function is cos(something).

Because , the Chain Rule gives:

= –sin(x3) ⋅ (x3)

= –sin(x3) ⋅ 3x2


Solution

so:

= e5x ⋅ (5x) = e5x ⋅ 5 = 5e5x

Practice


1.

2.

3.

4.

5.

6.

7.

8.

9.

10. h(x) = sin(πx)

11.

12.

13.

14.

15.

16.

17. y =

18. y = ln(x sin(x))

e xx e

22( ) +

f 1x 2 � sec110x2 � ex 2

y � xe2x

f 1u 2 �sin12u 2

u

g1x 2 �ex � e�x

2

f 1x 2 � ex � e2x � e3x

y � 1 ln1x 2 25

y � ln13t � 5 2

g x x( ) tan( )=

f x x( ) tan= ( )y � 23 ex � 1

g x x x( ) + + = 2 9 1

y � 1u5 � 3u4 � 7 272

h1t 2 � 1t8 � 9t3 � 3t � 2 210

y � 1x2 � 8x � 9 23

f 1x 2 � 18x3 � 7 24

d

dx

h¿ 1x 2 � e1something2 # ddx1something 2

h1x 2 � e1something2

h1x 2 � e5x

d

dx

dy

dx� �sin1something 2 # d

dx1something 2


83

“SOMETHING” HINT

It is important that the “something” in the parentheses appear somewhere in the derivative, just as it does inthe original function. If it doesn’t appear, then a mistake has been made.

tan(cos(x)) ≠ tan(x) ⋅ cos(x).

You CANNOT cancel the θ’s to conclude that sin(2θ)

θ = sin(2).

Calc2e_10_81-84.qxd 11/18/11 12:48 AM Page 83

19. s(u) =(sin(u) + cos(u))3

20. y = tan(cos(x))

This rule is called the Chain Rule because it works in

long succession when there are many layers to the

function. It helps to write out the function using lots

of parentheses, and then work patiently to take the

derivative of each outermost layer.

Example

Differentiate .

SolutionWith all of its parentheses, this function is

. The outermost layer is “some-

thing to the seventh power,” the second layer is “the

sine of something,” the third layer is “e raised to the

something,” and the last layer is 5x. Thus:

= 7(sin(e(5x)))6 ⋅ cos(e(5x)) ⋅ e(5x) ⋅ 5

= 35e(5x)sin6(e(5x))cos(e(5x))


Notice once again that every part except the out-

ermost layer (the natural logarithm) appears some-

where in the derivative.

Practice


21.

22.

23.

24.

25.

26. h xe

e

x

x( ) cos= −

−

2

4 2

11

k1u 2 � sec1 ln18u3 2 2

y � sin1sin1sin1x 2 2 2

g1t 2 � ln1tan1et � 1 2 2

y � 1e9x2�2x�1 24

f 1x 2 � cos318x 2

y � ln1x3 � tan13x2 � x 2 2

= ⋅ ⋅ ⋅7 55 6 5 5(sin( )) cos( ) ) ( )( ) ( ) ( )e e e xx x x d

dx

= ⋅ ⋅7 5 6 5 5(sin( )) cos( ) ( )( ) ( ) ( )e eddx

ex x x

f ¿ 1x 2 � 71sin1e15x2 2 26 # ddx1sin1e15x2 2 2

f 1x 2 � 1sin1e15x2 2 27

f 1x 2 � sin71e5x 2

–CHAIN RULE–

84

Solution

= + + ++ +

⋅3 3 6 1

3

2 2 2

3 2

x x x x

x x x

sec ( ) ( )

tan( )

13x2 � sec213x2 � x 2 # 16x � 1 2 2=+ +

⋅133 2x x xtan( )

3 3 32 2 2 2x x xd

dxx x+ + +

⋅ sec ( ) ( )=+ +

⋅133 2x x xtan( )

dy

dx x x x

d

dxx x x=

+ ++ +⋅1

33

3 23 2

tan( )( tan( ))

Calc2e_10_81-84.qxd 11/18/11 12:48 AM Page 84

LE

SS

ON

IMPLICIT DIFFERENTIATION

Acommon complaint about the Chain Rule is “I don’t know where to stop!” For example, why do we

use the Chain Rule for to get , but not for ,

which has ? The honest answer is that we could use the Chain Rule as follows:

When we get down to , we know we are done. The advantage to this way of thinking is that it

explains what really means. This isn’t merely a symbol that says “we took the derivative.” This is the result

of differentiating both sides of an equation.

dy

dx

ddx1x 2 � 1

� cos1x3 2 # 3x2 # 1 � cos1x3 2 # 3x2f ¿ 1x 2 � cos1x3 2 # ddx1x3 2 � cos1x3 2 # 3x2 # d

dx1x 2

g¿ 1x 2 � cos1x 2 # ddx1x 2 � cos1x 2 # 1 � cos1x 2

g¿ 1x 2 � cos1x 2

g1x 2 � sin1x 2f ¿ 1x 2 � cos1x3 2 # 3x2f 1x 2 � sin1x3 2

11

85

Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 85


SolutionStart with the equation.

Differentiate both sides of the equation.

Use .

Simplify.

Now if , then there is a relationship

between y and x. This relationship is given explicitly

because we know exactly what y is in terms of x. How-

ever, if the variables x and y are all mixed up on both

sides of the equals sign, then the relationship is said to

be implicit. The relationship is implied, but it is up to

us to figure out what the relationship between x and y

is explicitly. For example, the equation of the unit cir-

cle is:

There is a relationship between the values of x

and y, because what y can be depends on the value of

x. If x � 0, for instance, then y could be either 1 or �1.

We could take the implicit description of y in

and make it explicit by solving for y:

Solving for y is not always possible, though.

In fact, it rarely is. If our equation were ln(y) + cos(y)

= 3ex – x3, then we would not be able to solve for y.

Fortunately, we can still find the slope of the tan-

gent line, , without having to solve the original

equation for y. The trick is to use implicit differentia-

tion by taking the derivative of both sides and making

sure to include wherever the Chain Rule

dictates.

ExampleFind the slope of the tangent line to .


Differentiate both sides.

Use the Chain Rule everywhere.

Use and .

Solve for .

It might make you uneasy to have given in

terms of both x and y, but this is necessary. If we were

dy

dx

dy

dx�

�2x2y

� �xy

dy

dx

2x # 1 � 2y # dy

dx� 0

ddx1y 2 �

dy

dxddx1x 2 � 1

2x # ddx1x 2 � 2y # d

dx1y 2 � 0

ddx1x2 � y2 2 �

ddx11 2

x2 � y2 � 1

x2 � y2 � 1

ddx1y 2 �

dy

dx

dy

dx

y x 1 = ± − 2

y2 � 1 � x2

x2 � y2 � 1

x2 � y2 � 1

y � 4x5 � ex

dy

dx� 20x4 # 1 � ex # 1 � 20x4 � ex

dy

dx� 20x4 # d

dx1x 2 � ex # d

dx1x 2

ddx1y 2 �

dy

dx

ddx1y 2 �

ddx14x5 � ex 2

y � 4x5 � ex

y � 4x5 � ex

–IMPLICIT DIFFERENTIATION–

86

Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 86

asked, “What is the slope of the tangent line to

at ?” We would have to reply,

“Which one?” There are two tangent lines when

! See Figure 11.1. If we want the slope of

the tangent line at , then

.

Example

Find when .




Use and .

Factor out a .

Solve for .

To get rid of the fraction-in-a-fraction, we can

multiply the top and bottom by the denominator y

that we want to eliminate:

�3yex � 3x2y

1 � ysin1y 2

� a3ex � 3x2

1y � sin1y 2

b # a y

yb

dy

dx�

3ex � 3x2

1y � sin1y 2

dy

dx�

3ex � 3x2

1y � sin1y 2

dy

dx

a1y

� sin1y 2bdy

dx� 3ex � 3x2

dy

dx

1y

# dy

dx� sin1y 2 # dy

dx� 3ex � 3x2

ddx1y 2 �

dy

dxddx1x 2 � 1

3ex # ddx1x 2 � 3x2 # d

dx1x 2

1y

# ddx1y 2 � sin1y 2 # d

dx1y 2 �

ddx1 ln1y 2 � cos1y 2 2 �

ddx13ex � x3 2

ln1y 2 � cos1y 2 � 3ex � x3

ln1y 2 � cos1y 2 � 3ex � x3dy

dx

dy

dx

x

y = =

−= =

12

32

1

3

3

3

1

2

3

2, −

x �12

x �12

x2 � y2 � 1


87

1

1–1

y

x__2

x2 y2= 1+

)32–( 1– –

2,

1

–1

)32–( 1– –

2,

Figure 11.1

Don’t forget to apply the Chain Rule when com-

puting dx d (cos(y)) when y depends on x.

Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 87

ExampleFind the slope of the tangent line to

at (1,�5).



Use the product rule on .

Use and .

Plug in x � 1 and y � �5.

Use .

Thus, the slope of the tangent line at (1,�5) is 25.

Example

Find when cos(x ⋅ sin(y)).



Compute the derivatives on both sides.

Use (x) = 1 and .d

dxy

dy

dx( ) =

d

dx

sec ( ) ( )

sin sin( ) sin( ) cos( ) ( )

2 yddx

y

x y y yddx

y x

⋅

= − ⋅( )⋅ + ⋅

⋅

ddx

yddx

x ytan( ) cos sin( )( ) = ⋅( )( )

tan( ) cos sin( )y x y= ⋅( )

dy

dx

25 �dy

dx

ln11 2 � 0

2 5 11

15

0

2( ) ln( ) ( )− ⋅ ⋅ + ⋅ − ==

dy

dx

dy

dx{

2y # dy

dx# ln1x 2 �

1x

# y2 �dy

dx

ddx1y 2 �

dy

dxddx1x 2 � 1

2y # ddx1y 2 # ln1x 2 �

1x

# ddx1x 2 # y2 �

ddx1y 2 � 0

y2ln1x 2

ddx1y2ln1x 2 2 �

ddx1y � 5 2

y2ln1x 2 � y � 5

y2ln1x 2 � y � 5


88

Simplify the right-hand side.

sec ( )

sin sin( ) sin( ) sin sin( ) cos( )

2 ydydx

x y y x y y xdydx

⋅

= − ⋅( )⋅ − ⋅( ) ⋅ ⋅ ⋅

sec ( ) sin sin( ) sin( ) cos( )2 ydydx

x y y ydydx

x⋅ = − ⋅( )⋅ + ⋅

⋅

Bring both instances of to the same side.

Factor out a .

sin sin( ) sin( )x y y= − ⋅( )⋅

sec ( ) sin sin( ) cos( )2 y x y y xdydx

+ ⋅( ) ⋅ ⋅[ ] =

dy

dx

sec ( ) sin sin( ) cos( )

sin sin( ) sin( )

2 ydydx

x y y xdydx

x y y

⋅ + ⋅( ) ⋅ ⋅ ⋅

= − ⋅( )⋅

dy

dx

Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 88

Solve for .

ExampleUse implicit differentiation and the fact that

to prove that .

SolutionIf , then the derivative of ln(x) is .

Raise both sides as powers of e.

Since and are inverses, .

Differentiate both sides.

Use the Chain Rule.

Solve for .

Use .

So, .

Practice

For questions 1 through 14, compute .

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15. Find the tangent line slope of

at (�3,1).

16. Find the tangent line slope of

at (1,�2).x3 � y3 � 3y � x

y3 � x2 � y2 � 5y � 14

sec1y 2 � 9y � x3cos1y 2

xy

� xy � x � y

x2y � y4x4

1y � x2 24 � 10x

y x y y y− = − − −ln( ) 3 2 1

sin( ) sin( ) sin( )x y x y− = −

y x y = +

tan1y 2 � cos1x 2

e e e ex x y y+ = +2 2

y2 � x � 3x4 � 8y

y y x − = ln( )

sin1y 2 � 4x � 7

y3 � y � sin1x 2

1y � 1 23 � x4 � 8x

dy

dx

ddx

xx

ln( )( ) = 1

dy

dx�

1x

ey � eln1x2 � x

dy

dx�

1ey

dy

dx

ey # dy

dx� 1

ddx1ey 2 �

ddx1x 2

ey � x

eln1x2 � xexln1x 2

ey � eln1x2

y � ln1x 2

dy

dxy � ln1x 2

ddx1 ln1x 2 2 �

1x

ddx1ex 2 � ex

dydx

x y y

y x y y x=

− ⋅( )⋅+ ⋅( ) ⋅ ⋅

sin sin( ) sin( )

sec ( ) sin sin( ) cos( )2

dy

dx


89

. To see this, let a = 16 and b

= 9. The left-hand side is , but the

right-hand side is .16 4 9 + 3 = 7+ =

25 5 =

a b a b+ ≠ +

Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 89

17. Find the slope of the tangent line to

at (4,2).

18. Find the slope of the tangent line at

on the graph of .


at the point .


at (0,0).ln ln( )e e yx y+( ) = −2

1

2 6,π

sin1y 2 � x

cos sin( )π

2 2y x

=

2

2 4,π

ln13y � 5 2 � x � y2


90

. To see this, let A

= π and B = . The left-hand side is sin

= 1, but the right-hand side is sin(π) – sin= 0 – 1 = –1.

π2

π2

π2

s A B s A s Bin in( ) in( )( )− ≠ −

Calc2e_11_85-90.qxd 11/18/11 12:49 AM Page 90

LE

SS

ON

RELATED RATES

Sometimes, both variables x and y depend on a third variable t. An equation relating x and y is often

able to be determined geometrically. Once you have gotten the hang of implicit differentiation, it

should not be difficult to take the derivative of both sides with respect to the variable t. This enables

us to see how x and y vary with respect to time t. The only difference is that , , and

so on. Only can be simplified.

ExampleAssume x and y depend on some variable t. Differentiate with respect to t.


Us the Chain Rule to differentiate both sides with respect to t.

ddt1y2 � cos1x 2 2 �

ddt14x2y 2

y2 � cos1x 2 � 4x2y

y2 � cos1x 2 � 4x2y

ddt1t 2 � 1

ddt1y 2 �

dy

dtddt1x 2 �

dxdt

12

91

Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 91

Use and .

Example

Assume x and y depend on some variable t. Differen-

tiate ex + y = y 3 + with respect to t.


ex + y = y 3 +

Differentiate both sides with respect to t.


Use and .

The variables need not be x and y, and the variable

upon which they depend need not be t.

ExampleAssume A, B, and C depend on some variable r. Dif-

ferentiate .

Solution

Practice

Assume all variables depend on t. Differentiate with

respect to t.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11. S = 6e2

12. D x y= +2 2

A �12

bh

C � 2p r

A � 4p r2

V �43p r3

A2 � B2 � C2

z �25

x2 �25

y2 �3

5x

ln1y 2 � ex � x2y2

2x � 2y � 10x3 � 7x

y4 � 3x2 � cos1y 2

y � 1x3 � x � 1 25

ddr

A Bddr

AC

dAdr

BdBdr

dAdr

CdCdr

A

C

3 4

3 8

2

2

+( ) = +

⋅ + ⋅ =⋅ − ⋅

π

3 4 2A BAC

+ = + π

edx

dt

dy

dty

dy

dt x

dx

dtx ⋅ ⋅ ⋅= + + 3

1

2

2

ddt1y 2 �

dy

dtddt1x 2 �

dxdt

ed

dtx

d

dty y

d

dty

x

d

dtxx ⋅ ⋅ ⋅=( ) ( ) ( ) ( )+ +3

1

2

2

d

dte y

d

dty xx( ) ( )+ = + 3

x

x

2y # dy

dt� sin1x 2 # dx

dt� 8xy # dx

dt�

dy

dt# 4x2

ddt1y 2 �

dy

dtddt1x 2 �

dxdt

8x # ddt1x 2 # y �

ddt1y 2 # 4x2

2y # ddt1y 2 � sin1x 2 # d

dt1x 2 �

–RELATED RATES–

92

= 0 because π is a constant.ddr

( )π

Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 92

Just as is a rate, so are , ,

, and so on. Because t typically represents time,

represents how fast y is changing

over time. Thus, if A is a variable that represents an

area, represents how fast that area is increasing or

decreasing over time.

Differentiating an equation with respect to t

results in a new equation, which shows how the rates

of change of the variables are related. For example, the

area A and radius r of a circle are related by:

Differentiating both sides with respect to t gives:

If a circle is growing in size, this equation details how

the rate at which the radius is changing, ,

relates to the rate at which the area is growing, .

ExampleA rock thrown into a pond makes a circular ripple that

travels at 4 feet per second. How fast is the area of the

circle increasing when the circle has a radius of 12 feet?

SolutionWe know that for circles A = πr2, so that,

. And we know that the radius is

increasing at the rate of feet per second, so

when the radius is r � 12 feet, the area is increasing at:

square feet per second

ExampleA spherical balloon is inflated with 40 cubic inches of

air every second. When the radius is 12 inches, how fast

is the radius of the balloon increasing? (Hint: The vol-

ume of a sphere with radius r is .)

Solution

We know that the volume of the balloon is increasing

at the rate of . We want to know the

value of when r � 12 inches. Differentiating

with respect to t gives:

When we plug in and r � 12 in, we get:

The radius of the balloon is increasing at the very

slow rate of inches per second.5

72p� 0.022

drdt

=⋅

=40 5 in.

4 144 sec in.

72 secπ π

40in3

sec� 4p 112 in 22 # dr

dt

dVdt

� 40in3

sec

dVdt

� 4p r2 # drdt

V �43p r3

drdt

dVdt

� 40in3

sec

V � 43p r3

� 96p � 301.6

� 96pft2

sec

dAdt

� 2p 112 feet 2 # 4feet

second

drdt

� 4

dAdt

� 2p r # drdt

dAdt

drdt

dAdt

� 2p r # drdt

A � p r2

dAdt

dy

dt�

y-change

t-change

dAdt

dy

dtdxdt

dy

dx�

y-change

x-change

–RELATED RATES–

93

Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 93

ExampleSuppose the base of a triangle is increasing at a rate of

8 feet per minute while the height is decreasing by 1

foot every minute. How fast is the triangle’s area

changing when the height is 5 feet and the base is 20

feet?

SolutionIf we represent the length of the base by b, the height

of the triangle as h, and the area of the triangle as A,

then the formula that relates them all is . The

base is increasing at and the height is

changing at . The �1 implies that 1

foot is subtracted from the height every minute, that is,

the height is decreasing. We are trying to find ,

which is the rate of change in area. When we differen-

tiate the formula with respect to t, we get:

When we plug in all of our information, includ-

ing the h � 5 feet and b � 20 feet, we get:

Thus, at the exact instant when the height is 5 feet

and the base is 20 feet, the area of the triangle is

increasing at a rate of 10 square feet every minute.

ExampleA 20-foot ladder slides down a wall at the rate of 2 feet

per minute (see Figure 12.1). How fast is it sliding

along the ground when the ladder is 16 feet up the

wall?

Solution

Here, because the ladder is sliding

down the wall at 2 feet per minute. We want to know

, the rate at which the bottom of the ladder is mov-

ing away from the wall. The equation to use is the

Pythagorean theorem.

If we plug in y � 16 ft and ft/min, we get:

We still need to know what x is at the particular instant

that y � 16, and for this, we go back to the

Pythagorean theorem.

, so x � ;12x2 � 144

x2 � 116 22 � 120 22

2 2 16 2 0xdxdt

⋅ + ⋅( )⋅ −

=ftft

min

dy

dt� �2

2x # dxdt

� 2y # dy

dt� 0

ddt1x2 � y2 2 �

ddt1202 2

x2 � y2 � 202

dxdt

dy

dt� �2

ftmin

dAdt

= ⋅

⋅( ) + −

⋅ ⋅ ( )

= − =

12

8 5 112

20

20 10 102 2 2

ftft

ftft

ft ft ft

min min

min min min

dAdt

�12

# dbdt

# h �dhdt

# 12

b

A �12

bh

dAdt

dhdt

� �1ft

min

dbdt

� 8ft

min

A �12

bh

–RELATED RATES–

94

ladder

20 feet

wall

y

x ground

Figure 12.1

Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 94

Using x � 12 (a negative length here makes no sense),

we get:

At the moment that y � 16 ft, the ladder is sliding

along the ground at feet per minute.

In the previous example, it was okay to say that the

hypotenuse was 20 because the length of the ladder

didn’t change. However, if we replace y with 16 in the

equation before differentiating, we would have implied

that the height was fixed at 16 feet. Because the height

does change, it needs to be written as a variable, y. In

general, any quantity that varies needs to be represented

with a variable. Only after all derivatives have been com-

puted can the information for the given instant, like

, be substituted.

Practice

13. Suppose and .

What is when x � �1 and y � 2?

14. Suppose . What is when

, x � 3, and y � �2?

15. Let . If and ,

what is when L � 0 and I � 3?

16. Suppose , , and

. What is when A � 2, B � 2,

and C � 1?

17. Suppose . If I increases by 4 feet

per minute and R increases by 2 square feet every

minute, how fast is A changing when I � 20?

18. Suppose . Every hour, K

decreases by 2. How fast is R changing when K

� 3 and ?

19. The height of a triangle decreases by 2 feet every

minute while its base shrinks by 6 feet every

minute. How fast is the area changing when the

height is 15 feet and the base is 20 feet?

20. The surface area of a sphere with radius r is

. If the radius is decreasing by

2 inches every hour, how fast is the surface area

shrinking when the radius is 20 inches?

21. A circle increases in area by 20 square feet

every hour. How fast is the radius increasing

when the radius is 4 feet?

22. The volume of a cube grows by 1,200 cubic

inches every minute. How fast is each side

growing when each side is 10 inches?

23. The surface area of a cube is decreasing at a

rate of 2 square inches per second. How fast is

an edge shrinking at the instant when each side

is 40 inches? (Hint: The surface area of a cube

with edge e is S = 6e2.)

A � 4p r2

R �14

K3 �1R2 � 11

A � I2 � 6R

dBdt

dCdt

� �2

dAdt

� 8A3 � B2 � 4C2

dKdt

dIdt

� 4dLdt

� 5K � eL � L � I2

dxdt

� 8

dy

dtxy2 � x2 � 3

dxdt

dy

dt� 5y2 � 3y � 6 � 4x3

y � 16

83

2 12 2 16 2 0

8

3

⋅ ( ) ⋅ + ⋅( ) ⋅ −

=

=

ft ftft

ft

dxdt

dxdt

min

min

95

CHANGING VALUES HINT

It is important to use variables for all of the values that are changing. Only after all derivatives have been com-puted can they be replaced by numbers.

Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 95

24. The height of a triangle grows by 5 inches each

hour. The area is increasing by 100 square

inches each hour. How fast is the base of the

triangle increasing when the height is 20 inches

and the base is 12 inches?

25. One end of a 10-foot long board is lifted

straight off the ground at 1 foot per second

(see Figure 12.2). How fast will the other end

drag along the ground after 6 seconds?

26. A kite is 100 feet off the ground and moving

horizontally at 13 feet per second (see Figure

12.3). How quickly is the string being let out

when the string is 260 feet long?

–RELATED RATES–

96

1 ft__sec

?

board

10 ft

Figure 12.2

ftsec

100 ftstring

?

12

Figure 12.3

Calc2e_12_91-96.qxd 11/18/11 12:51 AM Page 96

LE

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ON

LIMITS AT INFINITY

This lesson will serve as a preparation for the graphing in the next lesson. Here, we will work on ways

to identify asymptotes from the formula of a rational function, a quotient of two polynomials.

We’ve encountered vertical asymptotes informally in Lesson 5. They are easy to recognize for

rational functions because they occur at precisely those x-values at which the denominator equals zero and

the numerator does NOT equal zero. If both top and bottom are zero when evaluated at an x-value, you get

a small unshaded circle on its graph at that point. For example, has vertical asymp-totes at x � �3 and x � 4.

Horizontal asymptotes take a bit more work to identify. The graph will flatten out like a horizontal line

if large values of x all have essentially the same y-value.

In the graph of , in Figure 13.1 for example, if x is bigger than 5, then y will be very close to y

� 1. Similarly, if x is a large negative number, the corresponding y-value will be close to zero. Horizontal

asymptotes are related to the limits as x gets really big. For given in the graph:

and

In such case, we say that y = 1 and y = 0 are horizontal asymptotes of f.

lim ( )x

f x→−∞

= 0lim ( )x

f x→∞

= 1

f˛˛1x 2

y � f˛ 1x 2

f˛ 1x 2 �13x � 2 2 1x � 1 2

1x � 3 2 1x � 4 2

13

97

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 97

These limits at infinity (and negative infinity) identify

what the ends of the graph do. For example, if

, then the graph of will typi-

cally look something like that in Figure 13.2. If

, then the graph of will look

like that in Figure 13.3.

Notice that the infinite limits say only what hap-

pens way off to the left and to the right. Other calcu-

lations must be done to know what happens in the

middle of the graph.

The general trick to evaluating an infinite limit is

to focus on the most powerful part of the function.

Take for example.

There are several terms being added in this function.

However, the most powerful part is the term .

When x gets big enough, like when x � 1,000,000, then

This clearly rounds to 2,000,000,000,000,000,000,

which is the value of 2x3 at x = 1,000,000. It is in this

sense that is called the most powerful part of the

function. As x gets big, is the only part that counts.2x3

2x3

� 1,999,899,999,989,995,000

100,000,000,000,000 � 10,000,000 � 5,000

� 2,000,000,000,000,000,000 �

2x3 � 100x2 � 10x � 5,000

2x3

lim ( , )x

x x x→∞

− − −2 100 10 5 0003 2

y � h1x 2lim ( )x

h x→−∞

= ∞

y � g1x 2lim ( )x

g x→∞

= 3

98

ASYMPTOTE HINT

Notice that the graph of y = f (x) crosses both horizontal asymptotes. Vertical asymptotes cannot be crossedbecause they are, by definition, not in the domain. Horizontal asymptotes can be crossed, as illustrated in thisexample. Think of “asymptote” as meaning “flattens out like a straight line” and not “a line not to be crossed.”

1

2

3

1 2 3 4 5 6

y

x7 8 9 10–1–2–3–4–5

–1

–2

–3

y = f(x)

Figure 13.1

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 98

As x gets huge, is clearly even larger, and is

twice that. Thus, as x goes to infinity, so does . Basi-

cally, the higher the exponent of x, the more powerful

it is. With that in mind, the rules for infinite limits of

rational functions are fairly simple:

� If the numerator is more powerful, the limit goes

to or � .� If the denominator is more powerful, the limit

goes to 0.� If the numerator and denominator are evenly

matched, the limit is formed by the coefficients of

the most powerful parts.

∞∞

2x3

2x3x3

lim( , ) limx x

x x x x→∞ →∞

− − − = = ∞2 100 10 5 000 23 2 3

99

1

2

3

y

–1

y = g(x)

4

5

1

2

3

y

x

–1

y = g(x)4

5

OR

levels out like y = 3

levels out like y = 3

toward ∞ toward ∞

x

Figure 13.2

y

x

y = h(x)

does notlevel out

toward –∞

Figure 13.3

RULES FOR INFINITE LIMITS

The rules for Infinite Limits of Rational Functions are as follows:� If the numerator is more powerful, the limit goes to ∞ or �∞.� If the denominator is more powerful, the limit goes to 0.� If the numerator and denominator are evenly matched, the limit is formed by the coefficients of the

most powerful parts.

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 99

Example

Evaluate .

SolutionThe most powerful part of the numerator is , and

in the denominator is . Thus:

This limit is zero because the numerator is overpow-

ered by the denominator. Also, as x gets really big,

gets really close to zero. For example,

when x = 1,000, then .

Example

Evaluate .

SolutionHere, the numerator and denominator are evenly

matched, with each having as its highest power

of x.

The limit is formed by the coefficients of the

most powerful parts: 3 in the numerator and �4 in the

denominator.

Example

Evaluate .

SolutionHere,

As x goes to infinity, also gets really large, but the

negative in the �5 reverses this and makes

approach negative infinity.

Practice

Evaluate the following infinite limits.

1.

2. limx

x x x

x x→−∞

+ ++ −

4 3

5 8 1

2

3

3 10

limx

x x

x→∞

+ −+

5 2

8 1

2

4

3 10

�5x8

x8

lim lim

lim

x x

x

x x

x

x

x

x

→∞ →∞

→∞

− +−

=−

= − = −∞

5 7

1

5

5

5

2

10

2

8

10 4

limx

x x

x→∞

− +−

5 7

1

5

2

10 4

lim( )( )

lim

lim

lim

x x

x

x

x xx x

x xx

xx

→ − → −

→ −

→ −

∞ ∞

∞

∞

+ −− +

= + −−

=−

=−

= −

3 2 51 2 1 2

3 2 51 4

34

34

34

2 2

2

2

2

x2

lim( )( )x

x xx x→−∞+ −

− +3 2 5

1 2 1 2

2

1 1

1 0000 001

x = =

,.

1x

lim lim limx x x

x

x x

x

x x→∞ →∞ →∞

−+ +

= − = − =1

3 2

10

2

3

2

3

x3

�x2

limx

x

x x→∞

−+ +1

3 2

2

3

100

GOING TO INFINITY

The whole concept of “going to infinity” might be a bit confusing. This really means “going toward infinity,”because infinity is not reachable. Just know that “going to infinity” means that we see what happens when weplug really large numbers into the function, and that “going to negative infinity” means that we see what hap-pens when we plug really large negative numbers into the function.

Make certain to fully expand the polynomials inthe top and bottom of a rational function beforeidentifying the dominating terms in each.

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 100

3.

4.

5.

6.

7.

8.

9.

10.

The infinite limits of and can be seen

from their graphs in Figure 13.4.

In general, as x goes to infinity, is more pow-

erful than x raised to any number. The natural loga-

rithm, however, goes to infinity slower than any power

of x. It may look as though is beginning to

level out toward a horizontal asymptote, but actually,

it will eventually surpass any height as it slowly goes up

to infinity.

In more complicated situations, we use L’Hôpi-

tal’s rule. This states that if the numerator and

denominator both go to infinity (positive or negative),

then the limit remains the same after taking the deriv-

ative of the top and the bottom.

y � ln1x 2

ex

lim ln( )x

x→∞

= ∞ limx

e x

→−∞= 0lim

xe x

→∞= ∞

ln1x 2ex

limx

t

t→∞

+− 0, 000

6, 000, 000

22

limx

x

x→−∞

−+

2

2

1

1

limx

x x x

x x→−∞

+ −+

4 2

2

3 8

2

+ 4

+ 1

lim( )

x

x x

x→ ∞+

−5 2

1

2

2

lim( )( )t

t tt t→ ∞− +

− +8 3 11

1 3 1 3

4 3

2 2

lim( )( )t

tt t t→ −∞

++ −

14 1

limx

x x

x→∞

− −+

10 3 100

2

3

5

limx

x

x→∞

+−

5 2

2 1

–LIMITS AT INFINITY–

101

1

2

3

–2–3 1–1

y

x

y = ex

1

2

3

–1

1 2 3

–2

–3

y

x

y = ln(x)

Figure 13.4

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 101

Example

Evaluate .

Solution

Since and , we can use

L’Hôpital’s Rule.

Note: The little H over the equals sign indicates that

L’Hôpital’s Rule has been used at that point of the

computation. Examples like this demonstrate how

goes to infinity even slower than x does.

Example

Evaluate .

SolutionHere, and ,

so we can use L’Hôpital’s Rule.

Here, we need to use L’Hôpital’s Rule several more

times:

Hx

xe= ∞→

= ∞ lim6

lim limx

H

x

e

x x

e

x

x x

→∞ →∞+ +=

+3 4 5 6 42

=+ +→∞

limx

e

x x

x

3 4 52

H

x

ddx

ddx

e

x x x

x

=+ + +

→∞lim

( )

( )3 22 5 2

limx

e

x x x

x

→∞ + + +3 22 5 2

lim( )x

x x x→∞

+ + + = ∞3 22 5 2limx

xe→ ∞

= ∞

limx

e

x x x

x

→∞ + + +3 22 5 2

ln1x 2

limln( )

lim(ln( ))

( )

lim lim

x

H

x

x x

x

x

d

dxx

d

dxx

xx

→∞ →∞

→∞ →∞

−=

−

=−

= − =

11

1

10

1

lim( )x

x→∞

− = −∞1lim ln( )x

x→∞

= ∞

limln( )

x

x

x→∞ −1

102

L’HOPITAL’S RULE

If the numerator and denominator both go to infinity (positive or negative), the limit remains the same after tak-ing the derivative of the top and bottom. Using notation,

if and lim ( )x

g x→±∞

= ± ∞ lim ( )x

f x→±∞

= ± ∞ lim( )( )

lim( )( )x x

f xg x

f xg x→± →±∞ ∞

= ′′

When applying L’Hôpital’s Rule, we differentiatetop and bottom separately and form the quotientof them. We do NOT apply the Quotient Rule.

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 102

This example shows how is more powerful than

. If the denominator had an , we’d have to use

L’Hôpital’s Rule 100 times, but in the end, would

drive everything to infinity.

Example

Evaluate .

SolutionThe limit is not infinite. So we can’t use

L’Hôpital’s Rule. The function is only powerful

when x goes to positive infinity. Instead, we use the old

“plug in” method.

Example

Evaluate .

SolutionThis has the same problem as the previous example.

No matter what x may be, will always be

between �1 and 1. Thus, and so

Because and , the

function is squeezed between them to zero as

well: . This is called the Squeeze Theo-

rem or the Sandwich Theorem because of the way

is squished between two curves, both going

to zero.

Practice


11.

12.

13.

14.

15.

16.

17.

18.

19.

20. limx

x x

e xx→−∞

+ +−

4 5 2

7

3 2

3

limx

x x

e xx→∞

+ +−

4 5 2

7

3 2

3

limcos( )

x

x

x→∞

limx x

xe→ −∞

limln( )

x x

x x

e→∞+

2

limx

x

x

ee→∞

++

2

3

32

limx

x

x l x→∞

+−

3 22

n( )

limx

x x

x→−∞

+ −+

2 5 10

4 2

limx

x

x→∞

+−

5

1

limln( )

ln( )x

x

x→∞ +

3

5

sin1x 2

x2

limsin( )

x

x

x→∞=

20

sin1x 2

x2

limx x→∞

=10

2limx x→∞

− =10

2

�1x2 �

sin1x 2

x2 �1x2

�1 � sin1x 2 � 1

sin1x 2

limsin( )

x

x

x→∞ 2

limx

e

x x

x

→−∞ + −= =

5 7 10

0

something not zero

ex

limx

e x

→−∞= 0

limx

e

x x

x

→−∞ + −5 7 1

ex

x100x3

ex


103

y

x

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 103

Sign Diagrams

In order to calculate the limits at vertical asymptotes,

it is necessary to know where the function is positive

and negative. The key is this: A continuous function

cannot switch between positive and negative without

being zero or undefined. Functions are zero when the

top is zero and the bottom is NOT zero, and undefined

where the denominator is zero. Mark all of these points

on a number line. Between these points, the function

must be entirely positive or negative. This can be

found by testing any point in each interval.

For example, consider .

This function is zero at x � 4 and undefined at both

x � �2 and x � 1. We mark these on a number line

(see Figure 13.5).

In between x � �2 and x � 1, the function is

either always positive or always negative. To find out

which it is, we test a point between �2 and 1, such as

0. Because is negative, the func-

tion is always negative between �2 and 1. Similarly,

we check a point between 1 and 4, such as

; a point after 4, such as

; and a point before �2, such

as . The sign diagram for this

function is shown in Figure 13.6.

This makes calculating the limits at the

vertical asymptotes very easy. Not only does

have vertical asymptotes at

x � �2 and x � 1, but the limits are:

As before, we can calculate the limits at infinity:

Note: We do not use a sign diagram when determining

horizontal asymptotes.

Thus, has a horizontal asymptote of y � 0.

With all of this, we begin to get a picture of

, which can be seen in

Figure 13.7.

f˛1x 2 �x � 4

1x � 2 2 11 � x 2

f˛1x 2

lim( )( )x

x

x x→−∞−

+ −=

4

2 10

lim( )( )x

x

x x

x

x xx→∞−

+ −= −

− − +=

→∞

lim

4

2 1

4

20

2

lim( )( )x

x

x x→ +

−+ −

= ∞1

4

2 1

lim( )( )x

x

x x→ −

−+ −

= − ∞1

4

2 1

lim( )( )x

x

x x→− +

−+ −

= − ∞2

4

2 1

lim( )( )x

x

x x→− −

−+ −

= ∞2

4

2 1

f˛1x 2 �x � 4

1x � 2 2 11 � x 2

f˛1�3 2 ��7

�114 2�

74

f˛15 2 �1

71�4 2� �

128

f˛12 2 ��2

41�1 2�

12

f˛10 2 ��4211 2

� �2

f˛1x 2 �x � 4

1x � 2 2 11 � x 2


104

–2 1 4

Figure 13.5

++ --

–2 1 4

f(x)

Figure 13.6

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 104

Notice that the horizontal asymptote y � 0 is

approached from above as , because is

always positive when x � �2. At the other end, the

asymptote is approached from below as

because the function is negative when x � 4.

We shall deal with graphing more thoroughly in

the next lesson.

Practice

For questions 21 through 26, determine all asymptotes,

vertical and horizontal, of the following functions.

Also, make a sign diagram for each.

21.

22.

23.

24.

25.

26. m xx

x x x( )

( )( )( )=

+ + +

6

2 29 1 2

j xx x

( )( )( )

= −+ +

1

1 52

k1x 2 �2x � 1

x2 � 4x � 3

h1x 2 �x2 � 11x � 3 22

g1x 2 �x � 3x2 � 4

f˛1x 2 �x � 2x � 4

x → ∞

f˛1x 2x → −∞


105

1

2

3

4

5

6

–1–2–3–4–5–6 1 2 3 4 5 6–1

–2

–3

–4

–5

–6

y

x

Figure 13.7

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 105


27.

28.

29.

30.

31.

32. lim( )( )( )x

x

x x x→− + + +2

6

2 29 1 2

lim( )( )x x x→ +−

−+ +5

1

1 52

limxS3�

x � 1

x2 � 4x � 3

limxS �3�

x2 � 11x � 3 22

limxS2�

x � 3x2 � 4

limxS4�

x � 2x � 4


106

Remember, x2 + a2 ≠ (x – a)(x + a).

Calc2e_13_97-106.qxd 11/18/11 12:52 AM Page 106

LE

SS

ON

USING CALCULUS TO GRAPH

Here is where everything comes together! We know how to find the domain, how to identify asymp-

totes, and how to plot points. With the help of the sign diagrams from the previous lesson, we shall

be able to tell where a function is increasing and decreasing, and where it is concave up and down.

Quite simply, where the derivative is positive, the function is increasing. The derivative gives the slope

of the tangent line at a point, and when this is positive, the function is heading upward, viewed from left to

right. When the derivative is negative, the function slopes downward and decreases.

When the second derivative is positive, the function is concave up. This is because the second deriva-

tive says how the first derivative is changing. If the second derivative is positive, then the slopes are increas-

ing. If the slopes, from left to right, increase from –2, to –1, to 0, to 1, to 2, and so on, then the graph must

curve like the one in Figure 14.1. In other words, the curve must be concave up.

Similarly, if the second derivative is negative, the function curves downward like the one in Figure 14.2

and is concave down.

14

107

Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 107

The concavity governs the shape of the graph,

depending on whether the function is increasing

or decreasing. If is increasing and concave up

(thus, both and are positive), then the

graph has the shape shown in Figure 14.3.

If is increasing and concave down (thus,

is positive and is negative), then the

graph has the shape shown in Figure 14.4.

If is decreasing and concave down (thus,

both and are negative), then the graph

has the shape shown in Figure 14.5.

If is decreasing and concave up (thus, f ′(x)

is negative and f ″(x) is positive), the graph has the

shape of the one in Figure 14.6.

ExampleGraph .

Solution

This function is defined everywhere and thus has no

vertical asymptotes. Because

and , there

are no horizontal asymptotes.

The derivative

is zero at x �

�5 and x � 1. To form the sign diagram, we test:

, f ′(0) = –15, and . Note:

These points were chosen arbitrarily. Any point less

than �5 will give the same information as the value

x � �6, for instance, and any point between �5

and �1 will give the same information as the value at

x � 0. Thus, the sign diagram for is shown in

Figure 14.7.

f ¿ 1x 2

f ¿ 12 2 � 21f ¿ 1�6 2 � 21

31x2 � 4x � 5 2 � 31x � 5 2 1x � 1 2

f ¿ 1x 2 � 3x2 � 12x � 15 �

lim ( )x

x x x→−∞

+ − + = −∞3 26 15 1010) = ∞

lim(x

x x x→∞

+ − +3 26 15

f˛1x 2 � x3 � 6x2 � 15x � 10

f˛1x 2

f – 1x 2f ¿ 1x 2f˛1x 2

f – 1x 2f ¿ 1x 2f˛1x 2

f – 1x 2f ¿ 1x 2f˛1x 2

f˛1x 2

–USING CALCULUS TO GRAPH–

108

slope = –2

slope = –1

slope = 0

slope = 1

slope = 2

Figure 14.1

Figure 14.2

increasing concave up

+ =

Figure 14.3

concave downincreasing

+ =

Figure 14.4

decreasing concave down

+ =

Figure 14.5

concave updecreasing+ =

Figure 14.6

Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 108

Because the function increases up to x � �5 and

then decreases immediately afterward, there is a local

maximum at x � �5. The corresponding y-value is

. Thus, (�5,110) is a local maxi-

mum. Similarly, because the graph goes down to x �

1 and then goes up afterward, there is a local minimum

at x � 1. The corresponding y-value is , so

(1,2) is a local minimum.

A guideline for identifying local minimum and

maximum points is shown in Figure 14.8.

The second derivative is 6x + 12 =

6(x + 2), which is zero at x � �2. If we test the sign at

x � �3 and x � 0, we get and

. Thus, the sign diagram for is as

shown in Figure 14.9.

f – 1x 2f – 10 2 � 12

f – 1�3 2 � �6

f – 1x 2 �

f˛11 2 � 2

y � f˛1�5 2 � 110

109

–5 1

increasing decreasing increasing

+–

f(x)

+f '(x)

Figure 14.7

INCREASING OR DECREASING

Remember, the sign of f ′(x) determines whether f (x) is increasing or decreasing.

Note: We use f ′(x) to see if the graph is increasing or decreasing, but f (x) to find the y-value at a point.

increasing decreasing

local maximum increasingdecreasing

local minimum

Figure 14.8

+–f (x) –2

concave down concave up

f "(x)

Figure 14.9

Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 109

Clearly x � �2 is a point of inflection, because

this is where the concavity switches from concave

down to concave up. The y-value of this point is

.

Before we draw the axes for the Cartesian plane,

we should consider the three interesting points we

have found: the local maximum at (�5,110), the local

minimum at (1,2), and the point of inflection at

(�2,56). If our x-axis runs from x � �10 to x � 10,

and our y-axis runs from 0 to 120, then all of these

points can be plotted on our graph (see Figure 14.10).

Example

Graph .

SolutionThe domain is . There is a vertical asymptote at x

� 2. The sign diagram for is shown in Figure 14.11.

Thus, and .

Because and ,limx

x

x→−∞+−

=

3

21lim

x

x

x→∞+−

=

3

21

limx

x

x→ +

+−

= ∞2

3

2

lim

x

x

x→ −

+−

= − ∞2

3

2

g1x 2

x � 2

g1x 2 �x � 3x � 2

f˛1�2 2 � 56


110

–2–3–4–5–6 1 2–1

y

x

10

20

30

40

50

60

70

80

90

100

110

120(–5,110)

(–2,56)

(1,2)

f(x) = x3 + 6x2 – 15x + 10

–7–8–9–10 3 4 5 6 7 8 9 10

–5 1

–2

increasing/decreasing

concavity

Figure 14.10

+-+g(x)

–3 2 above x-axisbelow x-axisabove x-axis

Figure 14.11

Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 110

there is a horizontal asymptote at y � 1, both to the

left and to the right. The derivative

has the sign

diagram shown in Figure 14.12.

The second derivative has the

sign diagram shown in Figure 14.13.

Because we have no points plotted at all, it makes

sense to pick one or two to the left and right of the ver-

tical asymptote at x � 2. At x � 1, , so

(1,�4) is a point. At x � 3, , so (3,6) is

another point. At x � �3, , so (�3,0) is

another nice point to know. Judging by these, it will

be useful to have both the x- and y-axes run from

�10 to 10.

To graph , it helps to start with the points

and the asymptotes as shown in Figure 14.14.

g1x 2

g1�3 2 � 0

g13 2 � 6

g11 2 � �4

g– 1x 2 �10

1x � 2 23

1 # 1x � 2 2 � 1 # 1x � 3 2

1x � 2 22�

�51x � 2 22

g¿ 1x 2 �


111

– –g'(x)

2

decreasing decreasing

Figure 14.12

g "(x)

2

+–


Figure 14.13

–2–3–4–5–6 1 2–1

y

x

1

2

3

4

5

6

7

8

9

10

–7–8–9–10 3 4 5 6 7 8 9 10–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

(1,–4)

(–3,0)y = 1

x = 2

(3,6)

Figure 14.14

Don’t automatically assume that the signs in a signdiagram will alternate. In fact, they don’t preciselywhen the number on the line comes from a factorraised to an even power, like (x – 3)2.

Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 111

Then we establish the shapes of the lines through

these points using the concavity and the intervals of

decrease (see Figure 14.15).

Example

Graph .

Solution

To start, . Thus,

has vertical asymptotes at x � 1 and x � �1. The

sign diagram for is shown in Figure 14.16.h1x 2

h1x 2

h1x 2 �x2 � 1x2 � 1

�x2 � 1

1x � 1 2 1x � 1 2

h1x 2 �x2 � 1x2 � 1


112

–2–3–4–5–6 1 2

y

x

1

2

3

5

6

7

8

9

10

–7–8–9–10 3 4 5 6 7 8 9 10–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

(3,6)

(1,–4)

(–3,0)

g(x) = _____x – 2

2

2

increasing/decreasing

concavity

–1

4

x + 3

Figure 14.15

Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 112

Note: can never be zero. The limits at the ver-

tical asymptotes are thus:

Because and ,

there is a horizontal asymptote at y � 1.

The derivative is as follows:

Its sign diagram is shown in Figure 14.17. This

indicates that there is a local maximum at x � 0. The

corresponding y-value is .

The second derivative is as follows:

.

The sign diagram is shown in Figure 14.18. It looks like

there ought to be points of inflection at x � �1 and x

� 1, but these are asymptotes not in the domain, so

there are no actual points where the concavity changes.

Before we graph the function, it will be useful to

have a few more points. When x � �2, then

and when x � 2, as

well. Thus, it will be useful to have the x- and y-axes

run from about �3 to 3. We start with just the points

and asymptotes (see Figure 14.19).

Then we add in the actual curves, guided by the

concavity and the intervals of increase and decrease

(see Figure 14.20).

y � h12 2 �53

y � h1�2 2 �53

12x2 � 41x � 1 231x � 1 23

�12x2 � 41x2 � 1 23

�

��41x2 � 1 2 � 2 # 2x1�4x 2

1x2 � 1 23

�41x2 � 1 22 � 21x2 � 1 2 # 2x1�4x 2

1x2 � 1 24h– 1x 2 �

y � h10 2 � �1

�4x1x � 1 221x � 1 22

2 1 2 1

1

2 2

2 2

x x x x

x

( ) ( )

( )

− − +−

=h¿ 1x 2 �

limx

x

x→−∞

+−

=2

2

1

11

lim

x

x

x→∞

+−

=2

2

1

11

limx

x

x→ +

+−

= ∞1

2

2

1

1

limx

x

x→ −+−

= − ∞1

2

2

1

1

limx

x

x→− +

+−

= − ∞1

2

2

1

1

limx

x

x→− −

+−

= ∞1

2

2

1

1

x2 � 1


113

+h(x)

–1 1 above x-axisbelow x-axisabove x-axis

+ –

Figure 14.16

–1

h'(x)

1

increasing decreasingincreasing

0

decreasing

h(x)

+ + – –

Figure 14.17

+ – +h"(x)

1concave down concave up

–1concave up

Figure 14.18

Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 113

Practice

Use the asymptotes, concavity, and intervals of increase

and decrease, and concavity to graph the following

functions.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10. f˛1x 2 �x

x2 � 1

j1x 2 �x2 � 1

x

k1x 2 �x

x2 � 1

h1x 2 �1

x2 � 9

g1x 2 �x

x � 2

f˛1x 2 � x4 � 8x3 � 5

k1x 2 � 3x � x3

h1x 2 � 2x3 � 3x2 � 36x � 5

g1x 2 � �4x � x2

f˛1x 2 � x2 � 30x � 10


114

1

2

3

–1

–2–3 1 2 3–1

y

x

–2

–3

h(x) = x2 + 1______

0 1–1

1–1

decreasing

concavity

x2 – 1

increasing

–2, 5—32, 5—3

Figure 14.20

2

3

–1

–2–3 1 2 3–1

y

x

–2

–3

y = 1

x = 1x = –1

1

(0,–1)

–2, 5—3 2, 5—3

Figure 14.19

Calc2e_14_107-114.qxd 11/18/11 12:53 AM Page 114

LE

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ON

OPTIMIZATION

Knowing the minimum and maximum points of a function is useful for graphing and even more for

solving real-life problems. Businesses want to maximize their profits, builders want to minimize their

costs, drivers want to minimize distances, and people want to get the most for their money. If we can

represent a situation with a function, then the derivative will help find optimal points.

If the derivative is zero or undefined at exactly one point, then this is very likely to be the optimal point.

The first derivative test states that if the function increases before that point and decreases afterward, it is max-

imal (see Figure 15.1). Similarly, if the function decreases before the point and increases afterward, then the

point is minimal.

The second derivative test states that if the second derivative is positive, then the function curves up, so

a point of slope zero must be a minimum (see Figure 15.2). Similarly, if the second derivative is negative, the

point of slope zero must be the highest point on the graph. Remember that we are assuming that only one

point has slope zero or an undefined derivative.

If there are several points of slope zero and the function has a closed interval for a domain, then plug

all the critical points (points of slope zero, points of undefined derivative, and the two endpoints of the inter-

val) into the original function. The point with the highest y-value will be the absolute maximum, and the one

with the smallest y-value will be the absolute minimum.

15

115

Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 115

ExampleA manager calculates that when x employees are work-

ing at the same time, the store makes a profit of

dollars each hour. If there

are ten employees and at least one must be working at

any given time, how many employees should be sched-

uled to maximize profit?

SolutionThis is an instance of a function defined on a closed

interval because limits the options for x.

The derivative of the profit function is

= = –3(x –

2)(x – 8). Thus, the derivative is zero at x � 2 and

at x � 8.

Because the function is defined on a closed inter-

val, we cannot use the first or second derivative tests.

Instead, we evaluate f(x) at each of our critical points.

These are the points of slope zero, x � 2 and x � 8, plus

the endpoints of the interval, namely x � 1 and x � 10.

These are evaluated as follows: ,

, , and . If the

manager wants to maximize the store profit, eight

employees should be scheduled at the same time,

because this will result in a maximal profit of $64

each hour.

ExampleA coffee shop owner calculates that if she sells cookies

at $p each, she will sell cookies each day. If it costs

her 20¢ to make each cookie, what price p will give her

the greatest profit?

SolutionProfit is computed as: Profit � Revenue � Costs. If she

charges $p per cookie, then she’ll make and sell

cookies each day. Thus, her revenue will be200p2

200p2

P110 2 � 20P18 2 � 64P12 2 � �44

P11 2 � �34

�31x2 � 10x � 16 2� 30x � 48 � 3x2

P¿ 1x 21 � x � 10

P1x 2 � 15x2 � 48x � x3

–OPTIMIZATION–

116


increasingdecreasing

slope = 0

slope = 0

MAX

MIN

Figure 15.1

slope = 0

slope = 0

MAX

MIN


Figure 15.2

Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 116

and her costs will be

= 40p2 . Therefore, her profit function is

. We limit this to p � 0.20

because the only optimal situation would be when the

cookies were sold for more than it cost to make them.

The derivative is , which is

zero when and therefore , so

either p � 0 or 0.40. Because p � 0 is not

in the domain, the only place where the derivative is

zero is at p � 0.40.

Using the first derivative test, we see that

and . So the sign dia-

gram for P ′ is as shown in Figure 15.3. Thus, the

absolute maximal profit occurs when the cookies are

sold at 40¢ each.

ExampleAt $1 per cup of coffee, a vendor sells 500 cups a day.

When the price is increased to $1.10, the vendor sells

only 480 cups. If every 1¢ increase in price reduces the

sales by two cups, what price per cup of coffee will

maximize income?

SolutionHere, the income is Income � Price � Cups Sold. So

if x � the number of pennies by which the price is

increased, then . This

simplifies to . And, the deriva-

tive is . This is zero only when

. The second derivative is

, which is always negative, so x � 75 is

maximal by the second derivative test. Thus, the max-

imal income will occur when the price is raised by x �

75¢ to $1.75 per cup.

′′ = −I x( ) .0 04

x �3

0.04� 75

′ = −I x x( ) .3 0 04

I x x x( ) .= + −500 3 0 02 2

I x x x( ) ( . ) ( )= + ⋅ −1 0 01 500 2

′ = −P ( . )0 50 160′ =P ( . )0 30 740

80200

�p �

80p2 � 200p380p3 �

200p2

′ = − +P pp p

( )200 80

2 3

P pp p

( ) = −200 402

a200p2 b # 10.20 2

a200p2 b # p �

200p

–OPTIMIZATION–

117

0.40

P�(p)


+ –

Figure 15.3

Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 117

ExampleA farmer wants to build a rectangular pen with 80 feet

of fencing. The pen will be built against the side of a

barn, so one side won’t need a fence. What dimensions

will maximize the area of the pen? See Figure 15.4.

SolutionThe area of the pen is . We can’t take the

derivative yet because there are two variables. We need

to use the additional information regarding how much

fencing exists; there are 80 feet of fencing. Because no

fencing will be required against the barn wall, the total

lengths of the fence will be , thus

. We can plug this into the formula for

area in order to obtain .

Now we have a function of one variable

. The derivative is .

This is zero only when y � 20. Using the second deriv-

ative test, . So, the curve is concave down

and the point y � 20 is the absolute maximum. The

corresponding x-value is

. Therefore, the pen with the maximal area will be

x � 40 feet wide (along the barn) and y � 20 feet out

from the barn wall.

ExampleA manufacturer needs to design a crate with a square

bottom and no top. It must hold exactly 32 cubic feet

of shredded paper. What dimensions will minimize the

material needed to make the crate (the surface area)?

See Figure 15.5.

SolutionWe want to minimize the surface area of the crate. The

surface area of the box consists of four sides, each of area

, plus the bottom, with an area of . Thus,

the surface area is . Again, we need to

reduce this to a formula with only one variable in

order to differentiate. We know that the volume must

be 32 cubic feet, so . Thus,

. When we plug this into the surface area func-

tion, we get:

.

So, we have a function of one variable .A xx

x( ) = +128 2

Surface Area = 4 432 1282

22 2xy x x

xx

xx+ =

+ = +

y �32x2

Volume � x2y � 32

Area � 4xy � x2

x # x � x2x # y

�40

x � 80 � 2y � 80 � 2120 2

′′ = −A y( ) 4

′ = −A y y( ) 80 4A y y y( ) = −80 2 2

Area � x # y � 180 � 2y 2 # y

x � 80 � 2y

y � x � y � 80

Area � x # y

–OPTIMIZATION–

118

y y

x

barn wall

(overhead view)

pen

Figure 15.4

y

x

x

Figure 15.5

Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 118

The derivative is:

.

which is zero when

or , so x � 4.

The second derivative is:

,

which is positive when x � 4. So, the curve is concave

up and the sole point of slope zero is the absolute min-

imum. Thus, the surface area of the crate will be min-

imized if x � 4 feet and feet.

Practice

1. Suppose a company makes a profit of P(x) =

dollars when it makes

and sells x � 0 items. How many items should

it make to maximize profit?

2. When 30 orange trees are planted on an acre,

each will produce 500 oranges a year. For every

additional orange tree planted, each tree will

produce 10 fewer oranges. How many trees

should be planted to maximize the yield?

3. An artist can sell 20 copies of a painting at

$100 each, but for each additional copy she

makes, the value of each painting will go down

by a dollar. Thus, if 22 copies are made, each

will sell for $98. How many copies should she

make to maximize her sales?

4. A garden has 200 pounds of watermelons

growing in it. Every day, the total amount of

watermelon increases by 5 pounds. At the same

time, the price per pound of watermelon goes

down by 1¢. If the current price is 90¢ per

pound, how much longer should the

watermelons grow in order to fetch the highest

price possible?

5. A farmer has 400 feet of fencing to make three

rectangular pens. What dimensions x and y will

maximize the total area?

6. Four rectangular pens will be built along a

river by using 150 feet of fencing. What

dimensions will maximize the area of the pens?

x x

, , − +1 000 5 000

1002

y �32x2 �

3242 � 2

′′ = +A xx

( ) 256

23

x3 � 64�128x2 � 2x � 0

′ = − +A xx

x( ) 128

22

–OPTIMIZATION–

119

y

x

y

x

river (no fence needed)

Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 119

7. The surface area of a can is Area = 2πr2 + 2πrh,

where the height is h and the radius is r. The

volume is Volume = πr2h. What dimensions

minimize the surface area of a can with volume

16π cubic inches?

8. A painter has enough paint to cover 600 square

feet of area. What is the largest square-bottom

box that could be painted (including the top,

bottom, and all sides)?

9. A box with a square bottom will be built to

contain 40,000 cubic feet of grain. The sides of

the box cost 10¢ per square foot to build, the

roof costs $1 per square foot to build, and the

bottom will cost $7 per square foot to build.

What dimensions will minimize the building

costs?

10. A printed page will have a total area of 96

square inches. The top and bottom margins

will be 1 inch each, and the left and right

margins will be 32 inches each. What overall

dimensions for the page will maximize the area

of the space inside the margins?

h

r

–OPTIMIZATION–

120

y

x

printed

area

1

1

in.

in.

in.2__3

in.2__3

Calc2e_15_115-120.qxd 11/18/11 12:55 AM Page 120

LE

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ON

THE INTEGRAL AND AREAS UNDER CURVES

A round the same time that many great mathematicians focused on figuring out the slopes of tangent

lines, other mathematicians were working on an entirely different problem. They wanted to be able

to compute the area underneath any curve , such as the one shown in Figure 16.1.y � f˛1x 2

16

121

x

y = f(x)

y

a b

What is this area?

Figure 16.1

Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 121

The curvy nature of the upper curve y = f(x)

presents a problem when finding area. But we know

how to find areas of rectangles. Therefore, the approach

we shall take is to approximate the region using better

and better rectangular staircases, as follows.

–THE INTEGRAL AND AREAS UNDER CURVES–

122

Area of rectangular staircase

Number of Rectangles Diagram (Dashed Portion of the Region)

1 Area =

Bad approximation of the area of

the original region.

2 Area =

Better approximation than with

1 rectangle, but not great.

4 Area =

Even better approximation.

+ ⋅ + ⋅f f( ) ( )4 1 5 1height base height base{ { { {

f f( ) ( )2 1 3 1height base height base{ { { {⋅ + ⋅

1 2 3 4 5 x

y

y = f(x)

f f( ) ( )3 2 5 2height base height base{ { { {⋅ + ⋅

1 3 5 x

y

y = f(x)

f ( )5 4height base{ {⋅

5 x

y

y = f(x)

1

Figure 16.2

If we keep using more rectangles to dissect the original

region, the rectangles gobble up more of the region,

and so the approximation to the actual area is better.

The number to which these approximate areas get

close as we do so is called the integral of f on [a,b], and

is denoted .

Note: The “elongated S” symbol (called the integral

sign), , is used because the process involves a sum.

The integrand f (x)dx resembles the form of what we

are adding, namely height × base. The dx portion

doesn’t actually enter into the computation, though.

Example

Evaluate the integral .

Solution

This represents the area between the curve ,

the x-axis, the line x � 0, and the line x � 4 (see Fig-

ure 16.3). This region happens to be a triangle with a

height of 2 and a base of 4. The area of the triangle is

. Thus, .�4

0

12

x dx � 41212 2 14 2 � 4

y �12

x

�4

0

12

x dx

�

x

y

1 2 3

4

5

1

2

3

4

y = x_21

Figure 16.3

� f(x)dxb

a

Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 122

If the region is below the x-axis, the area is

counted negatively. Therefore, really repre-

sents “the area between the curve , the x-axis,

x � a, and x � b, where area below the x-axis is

counted negatively.”

Example

Evaluate the integrals f(x)dx, f(x)dx, f(x)dx,

and f(x)dx where the graph of is shown

in Figure 16.4.

Solution

First, f(x)dx = 2(2) = 4 because this area is a square

above the x-axis (see Figure 16.5).

Next, f(x)d = 2(2) + . (1) . 2 = 5 because this area

is a square plus a triangle (see Figure 16.6).

For f(x)dx, we must calculate how much area is

above the x-axis and how much is below (see Figure

16.7).

6

1�

1

2

4

1�

3

1�

y � f˛1x 23

1�

6

1�

4

1�

3

1�

y � f˛1x 2


123

1

2

3

1 2 3 4

y

x

y = f(x)

Figure 16.5

1

2

3

1 2 3 4

y

x

y = f(x)

Figure 16.6

1

2

3

1 2 3 4 5 6

y

x7–1–2

–1

–2

–3

y = f(x)

–4

Figure 16.4

� f(x)dxb

a

Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 123

There are 5 units of area above the x-axis and 4 units

below, so f(x)dx = 5 – 4 = 1.

Finally, f(x)dx represents a rectangle of area

4 that is entirely below the x-axis. Thus, f(x)dx =

(–4)(1) = –4 (see Figure 16.8).

Practice

Evaluate the following integrals.

Use the following graph to solve practice problems 1,

2, and 3.

1. f(x)dx 2. f(x)dx 3. f(x)dx


5, and 6.

4. g(x)dx 5. g(x)dx 6. g(x)dx6

0�

6

4�

4

0�

2

0�

1

0�

2

1�6

1�

7

6�

6

1�


124

1

2

3

1 2 3 4 5

y

x

y = f(x)

5

1

2

3

1 2 3 4 6

y

x7–1–2

–1

–2

–3

y = f(x)

–4

5 square units of areaabove the x-axis

4 square units ofarea below the x-axis

Figure 16.7

1

2

3

1 2 3 4 5

6

y

x

7

–1–2–1

–2

–3

y = f(x)

–4

Figure 16.8

1

2

3

1 2 3 4 5

y

x

y = g(x)

6

Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 124


8, and 9.

7. h(t)dt 8. h(t)dt 9. h(t)dt

Use the following graph to solve practice problems 10

through 12.

10. k(x)dx 11. k(x)dx 12. k(x)dx

For questions 13 through 18, compute the integral.

13. (x + 2)dx 14. 2dx 15. (t – 3)dt

16. xdx 17. 2xdx 18. (2x – 2)dx

You might have noticed that:

f(x)dx + f(x)dx = f(x)dx

The area between a and c is the area from a to b plus

the area from b to c, assuming, of course, that

a � b � c (see Figure 16.9).

Similarly, f(x)dx – f(x)dx = f(x)dx.

We can use these to perform calculations, even when

the exact functions are unknown.

Example

If f(x)dx = 7 and f(x)dx = 15, then what is

f(x)dx?

Solution

f(x)dx = f(x)dx + f(x)dx

= 7 + 15 = 22

10

3�

5

3�

10

3�

10

3�

10

3�

5

3�

b

a�

c

b�

c

a�

c

a�

c

b�

b

a�

8

0�

6

1�

5

–2�

5

1�

4

1�

4

0�

5

4�

6

4�

7

0�

6

4�

4

–1�

6

–1�


125

x

y = f(x)

y

a b c

Figure 16.9

1

1 2 3 4 5t

y = h(t)

6–1–1

–2

1

1 2 3 4 5x

y = k(x)

6–1–1

–2

2

7

Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 125

Example

If g(x)dx = 38 and g(x)dx = –12, then what is

g(x)dx ?

Solution

g(x)dx = g(x)dx – g(x)dx

= 38 – (–12) = 50

Practice

For questions 19 through 21, suppose f(x)dx ,

f(x)dx , and f(x)dx . Evaluate the following.

19. f(x)dx 20. f(x)dx 21. f(x)dx

For questions 22 through 24, suppose g(t)dt = –3,

g(t)dt = 8, and g(t)dt = –10. Evaluate the

following.

22. g(t)dt 23. g(t)dt 24. g(t)dt

For questions 25 through 27, suppose h(x)dx = 20,

h(x)dx = 12, and h(x)dx = –5. Evaluate the

following.

25. h(x)dx 26. h(x)dx 27. h(x)dx

For questions 28 through 30, suppose j(x)dx = 2,

j(x)dx = 3, j(x)dx = –4, and j(x)dx = 12.

Evaluate the following.

28. j(x)dx 29. j(x)dx 30. j(x)dx2

–1�

2

0�

3

2�

3

–1�

2

1�

1

0�

0

–1�

11

10�

10

1�

11

1�

10

–2�

1

–2�

11

–2�

10

5�

10

1�

14

5�

5

1�

14

10�

14

1�

11

0�

11

6�

7

0�

11

6�

7

6�

6

0�

10

8�

10

0�

8

0�

8

0�

10

8�

10

0�


126

Calc2e_16_121-126.qxd 11/18/11 12:56 AM Page 126

LE

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THE FUNDAMENTALTHEOREM OF CALCULUS

Here comes the resounding climax of calculus. It would be best to read this lesson with some bombastic

orchestral music like that of Wagner or Orff! The initial question here is innocent enough: If we make

a function using that “area under a curve” stuff, what would its derivative be? To make this precise,

suppose that our curve is (see Figure 17.1). We use the variable t in order to save x for something

else later.

y � f1t 2

17

127

t

y = f(t)

y

Figure 17.1

Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 127

Now let our “area under the curve function” be

� the area under the curve between 0

and some point x. That is, . This

area is illustrated in Figure 17.2.

ExampleIf and , then what is g(3)?

Solution

� the area beneath the

curve from 0 to 3. The graph of f (t) � 2t

and the region are shown in Figure 17.3. This region

is a triangle with base 3 and height 6, so

.

Practice

For questions 1 through 6, suppose and

. Evaluate the following.

(Hint: The area of a trapezoid with bases b1 and b2 and

height h is .

1.

2.

3.

4.

5.

6. g15 2

g14 2

g13 2

g12 2

g11 2

g10 2

A h b b= +1

2 1 2( )

g1x 2 � �x

0

f˛1t 2 dt

f t t( ) = +3 1

g13 2 � �3

0

2t dt �1213 2 16 2 � 9

y � 2t

g13 2 � �3

0

f˛1t 2 dt � �3

0

2t dt

g1x 2 � �x

0

f˛1t 2 dtf˛1t 2 � 2t

g1x 2 � �x

0

f˛1t 2 dt

y � f˛1t 2g1x 2

–THE FUNDAMENTAL THEOREM OF CALCULUS–

128

t

y = f(t)

y

x

This area is g(x)

Figure 17.2

1

2

3

4

5

6

–11 2 3–1

y

t4 5

y = f(t) = 2t

(3,6)

Figure 17.3

Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 128

For questions 7 through 12, suppose and

. Evaluate the following.

7.

8.

9.

10.

11.

12.

Now that you are familiar with , we

can answer the next question: What is the derivative of

?

Begin with the definition of the derivative.

Use .

Use .

Now the integral represents the

skinny little area just to the right of point x (see Figure

17.4). This is almost a rectangle with a base of h and a

height of , so the integral is almost

. As h goes to zero, this approximation gets

better. Therefore,

� lim

˛

hS0 h # f˛1x 2

h� lim

˛

hS0 f˛1x 2 � f˛1x 2

g¿ 1x 2 � lim

hS0

�x�h

x

f˛1t 2 dt

h

h # f˛1x 2

�x�h

x

f˛1t 2 dtf˛1x 2

�x�h

x

f˛1t 2 dt

g¿ 1x 2 � limhS0

�x�h

x

f˛1t 2 dt

h

�c

a

f˛1t 2 dt � �b

a

f˛1t 2 dt � �c

b

f˛1t 2 dt

g¿ 1x 2 � limhS0

�x�h

0

f˛1t 2 dt � �x

0

f˛1t 2 dt

h

g1x 2 � �x

0

f˛1t 2 dt

g¿ 1x 2 � limhS0

g1x � h 2 � g1x 2

h

g1x 2

g1x 2 � �x

0

f˛1t 2 dt

g15 2

g14 2

g13 2

g12 2

g11 2

g10 2

g1x 2 � �x

0

f˛1t 2 dt

f˛1t 2 � 7


129

t

y = f(t)

y

x x + h

x

x + hf(t) dt

(x, f(x))

Figure 17.4

Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 129

What does this mean? It means that the deriva-

tive of the function , which represents “the area

under the curve from 0 to x,” is the very function

used to draw the curve. It came as an amazing surprise

to the world of mathematics that the process of find-

ing the slope of a tangent line and the process of find-

ing the area under a curve were such inverses. In order

to find the area under a curve , we need to

find a function whose derivative is .

We can use this to evaluate using the

Fundamental Theorem of Calculus.

For example, the derivative of is

. Thus, the area under between

x � 3 and x � 5 is

. This is exactly the area of the trapezoid

under the line between x � 3 and x � 5.

Example

The derivative of is . Use this

to evaluate .

SolutionBy the Fundamental Theorem of Calculus,

, where . Thus,

.

Even if we had drawn out the graph of ,

how would we have been able to guess that the area of

the two shaded curves add up to exactly three? This is

why the Fundamental Theorem of Calculus is so

powerful! (See Figure 17.5.)

y � x2

�1312 23 �

131�1 23 �

83

�13

� 3

�2

�1

x2 dx � g12 2 � g1�1 2

g¿ 1x 2 � f˛1x 2�b

a

f˛1x 2 dx � g1b 2 � g1a 2

�2

�1

x2 dx

g¿ 1x 2 � x2g1x 2 �13

x3

y � 2x

52 � 32 � 16

�5

3

2x dx � g15 2 � g13 2 �

f 1x 2 � 2xg¿ 1x 2 � 2x

g1x 2 � x2

�b

a

f˛1x 2 dx

f˛1x 2g1x 2

y � f˛1x 2

f˛1x 2

g1x 2

130

THE FUNDAMENTAL THEOREM OF CALCULUS

The Fundamental Theorem of Calculus can be written as follows:

f(x)dx = f(x)dx – f(x)dx = g(b) – g(a), where g′(x) = f (x)a

0�

b

0�

b

a�

1

2

3

–1

–2–3 1 2 3–1

y

x

4

y = x2

Figure 17.5

Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 130

Example

If , then . Use this to evaluate

4x3dx.

Solution

4x3dx = g(1) – g(–1)

The answer is zero because there is exactly as much

area above the x-axis (which counts positively) as there

is below the x-axis (which counts negatively).

Practice

For questions 13 through 16, use

to evaluate the following.

13. (2x + 1)dx

14. (2x + 1)dx

15. (2x + 1)dx

16. (2x + 1)dx

For questions 17 through 20, use to

evaluate the following.

17. dx

18. dx

19. dx

20. dx

For questions 21 through 24, use

to evaluate the following.

21. cos(x)dx

22. cos(x)dx

23. cos(x)dx

24. cos(x)dx2π

3π2

�

π

3π4

�

π2

π6

�

π4

0�

ddx

x x(sin( )) cos( )=

x100

0�

x9

4�

x4

0�

x1

0�

ddxa

23

x 32b � 2x

4

0�

6

2�

1

–3�

3

1�

ddx

x x x( )2 2 1+ = +

� 14 � 1�1 24 � 1 � 1 � 0

1

–1�

1

–1�

g¿ 1x 2 � 4x3g1x 2 � x4

131


Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 131

Calc2e_17_127-132.qxd 11/18/11 12:58 AM Page 132

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ON

ANTIDIFFERENTIATION

The Fundamental Theorem of Calculus shows that the area under the curve, , can be calculated

using a function whose derivative is . Symbolically,

Because of this, the symbol , without the limits of integration, is used to represent the opposite of

taking the derivative. An integral like is called a definite integral because it represents a definite area.

An integral like is called an indefinite integral because it represents another function.

Thus, means the antiderivative of or “the function whose derivative is .” For exam-

ple, asks “whose derivative is ?” This could be because . However, it could also

be because . In fact, because the derivative of a constant is zero, could be

plus any constant. Therefore, we write where c is any constant.�2x dx � x2 � cx2

�2x dxddx1x2 � 5 2 � 2xx2 � 5

ddx1x2 2 � 2xx22x�2x dx

f˛1x 2f˛1x 2� f˛1x 2 dx

� f˛1x 2 dx

�b

a

f˛1x 2 dx

�

�b

a

f˛1x 2 dx � �g1x 2�ba � g1b 2 � g1a 2

g¿ 1x 2 � f˛1x 2g1x 2�

b

a

f˛1x 2 dx

18

133

Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 133

We usually simply write with-

out actually saying that c stands for “some constant.” In

many ways, the “plus c” is the trademark of the indef-

inite integral because every problem that begins

with ends with .

If we are dealing with a definite integral like

, then it does not matter what constant we

use. For example:

The “plus c” will always cancel out in the subtraction.

As such, we simply use c � 0 and write:

Example

Use to

evaluate and

.

Solution

Because , we

know that:

Also,

=

=

The general process for finding antiderivatives of

powers is fairly simple. To take the derivative of

, we first multiply by the exponent 5, and

then we subtract one from the exponent. Thus,

.

To antidifferentiate , we must do the

exact opposite of this process. First, we add one to the

exponent, and then we divide the result by the new

exponent. Thus, . In

general, we write:

if n � �1�xnˇdx �

xn�1

n � 1� c

�5x4ˇdx �

5x4�1

4 � 1� c � x5 � c

�5x4ˇdx

f ¿ 1x 2 � 5x4

f˛1x 2 � x5

8 � 40 � 6 � 11 � 10 � 3 2 � 54 � 14 � 40

1 11 23 � 10 # 11 2 � 3 # 11 2 21 12 23 � 10 # 12 22 � 3 # 12 2 2 �

�2

1

13x2 � 20x � 3 2 dx � �x3 � 10x2 � 3x�21

� x3 � 10x2 � 3x � c� 13x2 � 20x � 3 2 dx

ddx1x3 � 10x2 � 3x 2 � 3x2 � 20x � 3

�2

1

13x2 � 20x � 3 2 dx

� 13x2 � 20x � 3 2 dx

ddx1x3 � 10x2 � 3x 2 � 3x2 � 20x � 3

�5

3

2x dx � �x2�53 � 52 � 32 � 25 � 9 � 16

= + − − =25 9 16 c c/ /

� 152 � c 2 � 132 � c 2

�5

3

2x dx � �x2 � c�53

�5

3

2x dx

� c� 1 p . 2 dx

�2x dx � x2 � c

134

BRACKET NOTE

The brackets are just a way of keeping track of the limits of integration a and b before they are pluggedinto g(x) and subtracted.

K[ ]ab

Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 134

Example

Evaluate x7dx.

Solution

Example

Evaluate

Solution

Example

Evaluate x3dx.

Solution

Practice


1. x 4dx

2. x12dx

3. u 6du

4. x5dx

5. xdx

6. t –3dt

ADDITION AND CONSTANT

�

9

–1�

1

–1��

�14

# 16 �14

# 0 � 4 � 0 � 4

= ⋅ − ⋅14

214

04 4

x dx x3 4

0

21

4=

2

0�

2

0�

�x

12 �1

12 � 1

� c �x

32

32

� c �23

x 3 2 � c

x dx12�xdx =�

xdx�

x dxx

c x c77 1

8

7 1

1

8

=+

+ = ++

�

�

135

VERIFICATION HINT

You can verify your answer by taking its derivative. If the derivative of your answer is what you were trying tointegrate, then you are correct.

The derivative of is . This verifies that

.xdx x c = +23

32�

ddx

x c x x x23

23

32

032

12

12 +

= ⋅ + = =23

32x c +

Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 135

RULES

g x dx( )b

a�f x dx( ) +

b

a�f x g x dx( ( ) ( )) + =

b

a�f x dx( ) ,

b

a�cf x dx c( ) =

b

a�

136

7. t–2dt

8.

9.

10.

11.

12.

Just as with derivatives, constants can stand aside,

and the terms of sums can be dealt with separately.

Example

Evaluate .

Solution

Example

Evaluate .

Solution

Example

Evaluate .

SolutionIt always helps to first write everything in exponential

form.

= + = 261

1283 329128,

= +

− +

4 82

2564

21

( )

= ⋅

−

−

= +

−6

23

84

42

32

4

1

4

32

4

1

4

xx

xx

6 8 512x x dx − −( )

4

1�6

85

xx

dx −

=4

1�

68

5x

xdx −

4

1�

214

812

7

12

4 7

4 2

4 2

t t t c

t t t c

= ⋅ − ⋅ + +

= − + +

7dt �tdt + �83t dt − �( )2 8 7 23t t dt− + = �

( )2 8 73t t dt− + �

513

53

2 3 3x dx x c x c= +

= + �5 52x dx =�

5 2x dx�

8dx4

–1�

5dx�

udu3�x dx4�

x dx53�

–1

–4�

Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 136

Practice


13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

The antiderivatives of , , and

follow directly from their derivatives:

because

because

because

The integral of will have to wait until Les-

son 20, though we can use the fact that

right now. We are inclined to say that

, but this is not entirely correct. The

derivative of is

as well. It does not matter if the

x inside the natural logarithm is positive or negative, so

we can generalize with the absolute value |x|.

Incidentally, this nicely fills a hole in an earlier

formula:

if n � �1x dxx

ncn

n

=+

++ 1

1�

1x

dx x c = +ln�

=−

− =⋅ 1

11

x x( )

�1

�x# d

dx1�x 2

ddx1 ln1�x 2 2ln1�x 2

x cln( ) +

1x dx =�

ddx1 ln1x 2 2 �

1x

ln1x 2

ddx1�cos1x 2 2 � sin1x 2

( )10 4 14u u du − +2

0�


cos( )x dx s x c in( ) = +�

ddx1ex 2 � exe dx e cx x = +�

cos1x 2sin1x 2ex

( )3 8103

47x x dx− �

12 xdx9

1�

( )10 4 14u u du − +2

0�

2 43 2t t t dt− −− −( )�

( )1 2 − t dt3

0�

( )3 911 2t t t dt+ + �

( )3 42x dx+ 2

1�

12 3x dx2

0�

( )6 10 52x x dx− + �x x dx3 −( )�

8 2u du�

9 4x dx�

–ANTIDIFFERENTIATION–

137

Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 137

and if then

Example

Evaluate .

Solution

Example

Evaluate .

Solution

Example

Evaluate .

Solution

Practice


25.

26.

27.

28.

29.

30.

31.

32. 8cos( )x dx5π2

π6

�

4e dxxln(3)

ln(2)�

( )x e dxx +1

0�

5 2sin( )x e dxx+( )�( sin( ))θ θ θ + 2 d�

2u du

e3

e�

( )3 2 3e x dxx + �( cos( ))x x dx2 5− �

= + + + − +

= + + + − +

−

13

12

13

12

1

3 2 1 1

3 2

x x x x x c

x x x xx

c

ln

ln

x x x x x dx2 1 1 2+ + + +( )− − 0�

x xx x

dx22

11 1+ + + +

= �

x xx x

dx22

11 1+ + + +

�

= − + = − 1 5 5 6 5e e

= − − − ( ) ( )1 5 0 53 1 3 0e e

( ) [ ]3 5 52 301t e dt t et t− = −

1

0�

( )3 52t e dtt−1

0�

�3cos1x 2 � 5sin1x 2 � c

( sin( ) cos( ))3 5x x dx + =�

( sin( ) cos( ))3 5x x dx +�

xdx x c= +1

ln�x dx− =1 �n � �1

–ANTIDIFFERENTIATION–

138

Calc2e_18_133-138.qxd 11/18/11 12:59 AM Page 138

LE

SS

ON

INTEGRATION BY SUBSTITUTION

The opposite of the Chain Rule is an integration technique called substitution. Using the Chain Rule,

for example, the derivative of is � �

. The corresponding antiderivative is thus � . It is easy

to recognize this after seeing the derivative worked out, but how should we know this otherwise?

The mantra of the Chain Rule is “multiply by the derivative of the inside.” So the first step to undoing

it is to identify what “the inside” must have been. We substitute a new variable u for this and then try to rewrite

the whole integral in terms of u.

For example, when confronted by , we first notice that if we multiplied out the

fourth power, then it would be a polynomial that we know how to evaluate, but doing so would be very tire-

some! Instead, we guess that “the inside” is the stuff inside the parentheses, and substitute .u � 3x2 � 7

�240x13x2 � 7 24 dx

813x2 � 7 25 � c�240x13x2 � 7 24 dx240 3 72 4x x( )+

8 # 513x2 � 7 24 # 6xddx1813x2 � 7 25 2813x2 � 7 25

19

139

Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 139

To convert the integral entirely in terms of u, we

must actually get a du into the integrand. Because u =

3x2 + 7, we know that = 6x, so du = 6xdx, or equi-

alently, dx = . So what? The steps below will illus-

trate why this is so useful.

Start with the original integral.

Substitute and .

Simplify.

Note, in particular, that substituting in dx =

resulted in the cancellation of all remaining x’s in the

integrand, so that the integral is not entirely in terms

of u.

Evaluate.

Replace .

Thus, � , as

we saw earlier.

In general, try using something inside parenthe-

ses with u. If every x doesn’t cancel out when replacing

dx with the expression involving du, then try using

something else as u. Sometimes, the entire denomina-

tor can be used as u. Sometimes, nothing works and a

different technique must be tried.

Example

Evaluate .

SolutionIf we use the stuff inside the only set of parentheses,

then , and thus and .


Substitute and .

Simplify.

Every x is gone, so we can evaluate.

Replace .

Thus, � . This can be

verified by differentiating �

� .x2sin1x3 2�131�sin1x3 2 # 3x2 2 � 0

ddx

x c− +

13

3cos( )

�13

cos1x3 2 � c�x2sin1x3 2 dx

�13

cos1x3 2 � c

u � x3

�13

cos1u 2 � c

� 13

sin1u 2 du

�x2sin1u 2 du3x2

dx �du3x2u � x3

�x2sin1x3 2 dx

dx �du3x2du � 3x2

dxu � x3

�x2sin1x3 2 dx

813x2 � 7 25 � c�240x13x2 � 7 24 dx

813x2 � 7 25 � c

u � 3x2 � 7

8u5 � c

du

x6

�40u4 du

�240x1u 24 du6x

dx �du6x

u � 3x2 � 7

�240x13x2 � 7 24 dx

du

x6

du

dx

–INTEGRATION BY SUBSTITUTION–

140

Before actually computing the integral, all of thex’s must be gone! You cannot integrate with mixedvariables.

Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 140

If we had been faced with in the last

example, then substituting would have

resulted in . This cannot be evaluated

because it is not entirely in terms of u. In fact, this inte-

gral is very difficult to solve and requires the advanced

technique of replacing with an infinitely long

polynomial called a power series. Many such integrals

exist that are difficult to solve. This book will focus on

the ones that can be evaluated with basic techniques.

Example

Evaluate .

SolutionBecause there are no parentheses, try using the

denominator: . Here, , so

and .


Substitute and .

Simplify.


Replace .

Thus, � .

Basically, the goal is to find a u whose derivative

is essentially the rest of the integral, except for possibly

a constant, so that between the u and the du, every x

goes away. This leads to some clever tricks, as will be

demonstrated in the following examples.

Example

Evaluate .

SolutionHere, we use . This is not because it is in

parentheses but because its derivative makes

up the rest of the integral. Here, , so

.


Substitute and .

�ux

1x du 2

dx � x duu � ln1x 2

� ln1x 2

x dx

dx � x du

du �1x

dx

dudx

�1x

u � ln1x 2

� ln1x 2

x dx

32

ln|2x � 7| � c� 32x � 7

dx

32

ln|2x � 7| � c

u � 2x � 7

32

ln|u| � c

� 32

# 1u

du

� 3u

du2

dx �du2

u � 2x � 7

� 32x � 7

dx

dx �du2

du � 2 dx

dudx

� 2u � 2x � 7

� 32x � 7

dx

sin1x3 2

�sin1u 2 du3x2

u � x3

�sin1x3 2 dx


141

Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 141

Simplify.


Replace .

Thus, � .

Example

Evaluate .

Solution

Here, the trick is to use so that

and .


Substitute and .

Simplify.


Replace .

Thus, � .

To use substitution on a definite integral, evalu-

ate the indefinite integral first, and then compute at the

limits.

Example

Evaluate .

Solution

First, evaluate the following indefinite integral.

Use u = x3, du = 3x2dx, and so .

Simplify.

Every x is gone, so we can evaluate the indefinite integral.

Replace u = x3.

Now, because , it follows that

x e dx e e e ex x2 3 3

0

113

13

13

13

11 0=

= − = −( )1

0�

x e dx e cx x2 3 313

= +�

13

3e cx +

13

e cu +

13

e duu�

x edux

u223

�dx

dux

=3 2

x e dxx2 3�

x e dxx2 31

0�

�14

cos41x 2 � csin( )cos ( )x x dx3�

�14

cos41x 2 � c

u � cos1x 2

�14

u4 � c

−u du3�

sin( )sin( )

x udu

x ⋅ −

3�

dx � �du

sin1x 2u � cos1x 2

sin( )cos ( )x x dx3�

dx � �du

sin1x 2dudx

� �sin1x 2

u � cos1x 2

sin( )cos ( )x x dx3�

121 ln1x 2 22 � c

ln( )x

xdx�

121 ln1x 2 22 � c

u � ln1x 2

12

u2 � c

udu�


142

Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 142

You will know that you chose the wrong u if

either some of the variables x still remain or the sim-

plified integral will still be hard to solve. If this happens,

go back to the beginning and try a different u. Don’t

forget that many integrals, like those of the previous

lesson, don’t require substitution at all. Like much of

mathematics, learning to integrate often requires

patience and a knack that is developed with practice.

Practice


1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24. ee

dxx

x

2

21+�

sin( )

cos( )

x

xdx�tan( )x dx =�

ex

dxx�

1

x xdx

ln( )�

(ln( ))x

xdx

3

�

e e dxx xsin( )�

sin( )7 2x dx −�

4cos( )x dx�

cos( )4x dx�

sin ( )cos( )2 x x dx�

sin( )cos( )x x dx�

−+2

4 10xdx

�

x

xdx

( )4 52 3+ �

( )( )8 5 4 5 12 3x x x dx + + −�

6 1

3 2 1

3

4

x

x xdx

−

− +

�

2 3 4x x dxcos( )�

9 5

3 5

2

3

x

xdx

−−

�

15 − x dx1

0�

2 1x dx+�

x x dx3 23 1⋅ − �

( )x x dx3 9 4− + �

x x dx2 3 41( )− 1

0�

( )4 34 10x dx+ �

x x dx4 5 71( )+ �


143

Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 143

Calc2e_19_139-144.qxd 11/18/11 1:00 AM Page 144

LE

SS

ON

INTEGRATION BY PARTS

The integral of the product of two functions is unfortunately not the product of the integrals. For

example, the integral is not ( ) .( ) . We

know this because the derivative of is, by the Product Rule, �

, which is not equal to . It is unfortunate that this does not work because,

if it did, evaluating integrals would be simple and would not require so many different techniques.

The integration technique that undoes the Product Rule is called integration by parts. We derive the for-

mula as follows.

The product rule for differentiating f(x) ⋅ g(x) says

.

Integrating both sides of the formula gives .

This simplifies to .g x f x dx( ) ( )′ ⋅�f x g x dx( ) ( )′ ⋅ + �f x g x( ) ( )⋅ =

f x g x g x f x dx( ) ( ) ( ) ( )′ ⋅ + ′ ⋅[ ]�ddx

f x g x dx( ) ( )⋅( ) =�

ddx

f x g x f x g x g x f x( ) ( ) ( ) ( ) ( ) ( )⋅( ) = ′ ⋅ + ′ ⋅

x # cos1x 2x x x x⋅ + ⋅sin( ) cos( )12

2

ddxa

12

x2sin1x 2 � cb12

x2sin1x 2 � c

x x c=

⋅( ) +sin( )1

22x dxcos( )�x dx�x x dx ⋅ cos( )�

20

145

Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 145

–INTEGRATION BY PARTS–

146

Now, we introduce two variables, u and v, as follows to simplify the formula:

Plug these into the above formula to get preceding.

Simply put, uv = vdu + udv. Move the first integral on the right-hand side to the left and voilà!, we have

the integration by parts formula:

vdu.�udv uv= −�

��

f x g x dxu dv

( ) ( ){ 124 34⋅ ′�g x f x dx

v du

( ) ( ){ 124 34⋅ ′ +�f x g x

u v

( ) ( ){ {⋅ =

u f xdu

dxf x du f x dx

v g xdv

dxg x dv g x dx

= = ′ = ′

= = ′ = ′

( ) ( ), ( )

( ) ( ), ( )

so that which is equivalent to

so that which is equivalent to

Work through the following examples to see how this

is applied to help us compute more complicated inte-

grals.

This can often be used to transform a difficult

integral into one that is solvable. For example, take

. This looks just like udv if and

. In order to use the formula, we

will need to get by differentiating u. Because

u = x, we know that , so . We will

also need to get v from dv by integrating. And because

dv = cos(x)dx, it must be that . Thus:

�

This is the correct answer, as can be verified by taking

the derivative � 1 ⋅ sin(x) +

cos(x) ⋅ x – sin(x) + 0 = x ⋅ cos(x).

Example

Evaluate .

SolutionThis cannot be solved by basic integration or by sub-

stitution. Since it is a product, there is a good chance

that integration by parts will work. First, try u = x. The

dv must then be everything else after the integral sign,

so . After differentiating u and integrating

dv, we get:

And:

Note: Although there is technically a “+ c ” here, we will

wait until the end to add the “+ c” once all integrals

have been computed.

v � ex

dv � ex dx

du � dx

u � x

dv � ex dx

xe dxx�

ddx1xsin1x 2 � cos1x 2 � c 2

xsin1x 2 � cos1x 2 � c

x dxsin( )�= − sin( ) x x

vdu�= − uv

udv �x x dx ⋅ =cos( )�v � sin1x 2

du � dxdudx

� 1

du

dv � cos1x 2 dx

u � x�x x dx ⋅ cos( )�

The dx must be part of what you decide to letequal dv.

Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 146

Thus, using the integration by parts formula

, we evaluate as follows:

�

�

�

�

Example

Evaluate .

SolutionHere, since we don’t know how to integrate ln(x)

(yet!), we can’t let it equal dv. So, we set and

. Thus:

And:

And then we evaluate.

�

�

�

�

�

Example

Evaluate .

SolutionBecause there seems to be only one part to this inte-

gral, one wouldn’t think to try integration by parts

first. However, because nothing else will work, we can

try . The only thing left for the dv is dx, so

we use dv � dx, which leads to v � x.

And:

dv � dx

v � x

And now evaluate as follows.

�

�

�

�

�

Sometimes, integration by parts needs to be done

more than once to compute an integral.

xln1x 2 � x � c

xln1x 2 � �1 dx

ln1x 2 # x � �x # 1x

dx

uv � �v du

�u dv� ln1x 2 dx

du �1x

dx

u � ln1x 2

u � ln1x 2

� ln1x 2 dx

14

x4ln1x 2 �1

16x4 � c

14

x4ln1x 2 � � 14

x3 dx

ln1x 2 # 14

x4 � � 14

x4 # 1x

dx

uv � �v du

�u dv�x3ln1x 2 dx

v �14

x4

dv � x3 dx

du �1x

dx

u � ln1x 2

dv � x3 dx

u � ln1x 2

�x3ln1x 2 dx

xex � ex � c

xex � �ex dx

uv � �v du

�u dv�xex dx

�u dv � uv � �v du


147

Don’t forget that there is a minus in the formula!

Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 147

Example

Evaluate .

Solution

Here, , so , and , so

.

�

�

�

�

In order to compute , we have to

use integration by parts a second time, but this time,

with and .

And:

Now we evaluate as follows.

�

�

�

�

�

Thus,

�

The final example utilizes a clever trick.

Example

Evaluate .

SolutionThe terms ex and sin(x) are equally good candidates for

u. Let us use

u = ex and dv = sin(x)dx.

And:

v � �cos1x 2

dv � sin1x 2 dx

du � exˇdx

u � ex

�exsin1x 2 dx

x2sin1x 2 � 2xcos1x 2 � 2sin1x 2 � c

�x2cos1x 2 dx

x2sin1x 2 � 2xcos1x 2 � ��2cos1x 2 dx

x2sin1x 2 � 12x # 1�cos1x 2 2 � � 1�cos1x 2 2 # 2 dx 2

x2sin1x 2 � 1uv � �v du 2

x2sin1x 2 � �u dv

x2sin1x 2 � �2xsin1x 2 dx�x2cos1x 2 dx

v � �cos1x 2

dv � sin1x 2 dx

du � 2 dx

u � 2x

dv � sin1x 2 dxu � 2x

�2xsin1x 2 dx

x2sin1x 2 � �2xsin1x 2 dx

x2sin1x 2 � �sin1x 2 # 2x dx

uv � �v du

�u dv�x2cos1x 2 dx

v � sin1x 2

dv � cos1x 2 dxdu � 2x dxu � x2

�x2cos1x 2 dx


148

Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 148

Now, evaluate as follows.

�

�

�

�

To evaluate , we use integration by parts

again, but with u = ex and dv = cos(x)dx.

And:

And then the evaluation:

�

�

�

�

Thus, we have:

�

Here is the moment of despair: To evaluate

, we need to be able to evaluate

! And yet, the trick here is to bring both

integrals to one side of the equation:

� �

�

�

= − +( ) +1

2e x x cx cos( ) sin( )

121�excos1x 2 � exsin1x 2 2 � c�exsin1x 2 dx

�excos1x 2 � exsin1x 22�exsin1x 2 dx

�excos1x 2 � exsin1x 2

�exsin1x 2 dx�exsin1x 2 dx

�exsin1x 2 dx

�exsin1x 2 dx

�excos1x 2 � exsin1x 2 � �exsin1x 2 dx

�exsin1x 2 dx

�excos1x 2 � exsin1x 2 � �sin1x 2 # ex dx

�excos1x 2 � uv � �v du

�excos1x 2 � �u dv

�excos1x 2 � �excos1x 2 dx�exsin1x 2 dx

v � sin1x 2

dv � cos1x 2 dx

du � exˇdx

u � ex

�excos1x 2 dx

�excos1x 2 � �excos1x 2 dx

ex1�cos1x 2 2 � � 1�cos1x 2 2 # ex dx

uv � �v du

�u dv�exsin1x 2 dx


149

Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 149

Practice

Evaluate the following integrals using integration by

parts, substitution, or basic integration.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22. e x dxx cos( )�

cos( ) ln sin( )x x dx⋅ ( )�

sin( ) cos( )x x dx�

e

xdx

x1

2�

e x dxx− cos( ) sin( )�

13x

dx�

15

x xdx

ln( )( )�

cos ( ) sin( )3 x x dx⋅�

xe dxx−�

x x dx − 1�

x dx − 1�

1x

x dx+

ln( )�

( )ln( )x x x dx3 3 1+ − �

ee

dxx

x

3

3 9+�

x e dxx2 13 + �

( sin( ))x x dx2 + �

x x dx2 sin( )�

ln( )x

xdx�

( )cos( )x x dx + 3�

x x dxsin( )2�

x x dxsin( )�

x x dx5 ln( )�


150

Calc2e_20_145-150.qxd 11/18/11 1:02 AM Page 150

If you have completed all 20 lessons in this book, you are ready to take the posttest to measure your progress.

The posttest has 50 multiple-choice questions covering the topics you studied in this book. Although the

format of the posttest is similar to that of the pretest, the questions are different.

Take as much time as you need to complete the posttest. When you are finished, check your answers with

the answer key that follows the posttest. Along with each answer is a number that tells you which lesson of

this book teaches you about the calculus skills needed for that question. Once you know your score on the

posttest, compare the results with the pretest. If you scored better on the posttest than you did on the pretest,

congratulations! You have profited from your hard work. At this point, you should look at the questions you

missed, if any. Do you know why you missed the question, or do you need to go back to the lesson and review

the concept?

If your score on the posttest doesn’t show much improvement, take a second look at the questions you

missed. Did you miss a question because of an error you made? If you can figure out why you missed the prob-

lem, then you understand the concept and simply need to concentrate more on accuracy when taking a test.

If you missed a question because you did not know how to work the problem, go back to the lesson and spend

more time working that type of problem. Take the time to understand basic calculus thoroughly. You need a

solid foundation in basic calculus if you plan to use this information or progress to a higher level. Whatever

your score on this posttest, keep this book for review and future reference.

POSTTEST

151

Calc2e_21_151-164_Post.qxd 11/18/11 1:03 AM Page 151

–LEARNINGEXPRESS ANSWER SHEET–

153

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

1.2.3.4.5.6.7.8.9.

10.11.12.13.14.15.16.17.

18.19.20.21.22.23.24.25.26.27.28.29.30.31.32.33.34.

35.36.37.38.39.40.41.42.43.44.45.46.47.48.49.50.

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d

a b c d


155

Posttest

1. Evaluate when .

a. �12

b. �10

c. �4

d. 4

2. Simplify when .

a.

b.

c.

d.

3. Evaluate (g ° h)(x) when

and .

a.

b.

c.

d.

4. What is the domain of ?

a. all real numbers greater than or equal to –1

b. all real numbers except 0

c. all real numbers except –1 and 0

d. all nonzero real numbers greater than or

equal to –1

Use the following graph for problems 5 through 7.

5. Where does have a local maximum?

a. x � 0

b. x � 1

c. x � 2

d. x � 3

6. On what interval(s) is decreasing?

a. (1,2) and (2,3)

b. (�q,2)

c. (�q,0)

d. (q,1) and (3,q)

g1x 2

g1x 2

f xx

x( )

= + 1

1x2 �

5x

� 1

1x2 � 5x � 1

x � 5 �1x

x xx

2 5 11+ + +

h1x 2 �1x

g1x 2 � x2 � 5x � 1

2x3 � 3x2 � x

2x2 � 3x

4x2 � 2x � 2

4x2 � 6x � 2

f˛1x 2 � x2 � xf˛12x � 1 2

f˛1x 2 � x3 � 2xf˛1�2 2

–POSTTEST–

1

2

3

–1–2 1 2 3 4–1

–2

–3

y

x

y = g(x)


–POSTTEST–

156

7. What is the slope of the line passing through

(2,�4) and (1,7)?

a. –

b. –11

c. 11

d. 3

8. Simplify .

a. 7

b. 12

c. 16

d. 64

9. Simplify .

a. �8

b. 4

c.

d.

10.Solve for x when .

a.

b. 2.5

c.

d.

11. Evaluate .

a. 1

b.

c.

d.

12. Evaluate .

a. –

b.

c. –

d.

13. Evaluate .

a. 0

b. 1

c. –2

d. undefined

limxS2

x2 � 5x � 6x2 � 2x � 3

2

2

2

2

3

2

3

2

sin a4p3b

3

2

2

2

12

cos ap

4b

ln a52b

104

ln110 2

ln14 2

4x � 10

116

14

16�12

43

111


14. Evaluate .

a. –2

b.

c. 1

d. undefined

15. Evaluate .

a. 5

b. �qc. qd. –4

16. What is the slope of the tangent line to f(x) = x2

at x � 3?

a. 2

b. 6

c. 9

d. 2x


at x � 2?

a. �7

b. �3

c. 1

d. 4

18. What is the derivative of

?

a. g′(x) = 0

b. g′(x) = – 1

c. g′(x) =

d. g′(x) =

19. Suppose that after t seconds, a falling rock is

feet off the ground.

How fast is the rock traveling after 2 seconds?

a. 10 feet per second

b. –32 feet per second

c. 59 feet per second

d. 146 feet per second

20. Differentiate y = + 4sin(x).

a.

b. + 4cos(x)

c. + 4sin(x)

d.1

24

xx − cos( )

dydx

=

x

2

dydx

=

xdydx

=

1

24

xx + cos( )

dydx

=

x

s1t 2 � �16t2 � 5t � 200

32x3 � 30x2 � 3

32x3 � 10x2 � 3x � x

8x4 � 10x3 � 3x

g1x 2 � 8x4 � 10x3 � 3x � 1

y � 4x � 7

limx

xx→ +

+−1

412

14

limxS3

x2 � 5x � 6x2 � 2x � 3

–POSTTEST–

157



a. f ′(x) =

b. f ′(x) =

c. f ′(x) =

d. f ′(x) =

22. Differentiate .

a.

b.

c.

d.

23. Differentiate .

a. g′(x) =

b. g′(x) =

c. g′(x) =

d. g′(x) =


a. f ′(x) =

b. f ′(x) =

c. f ′(x) =

d. f ′(x) = 0

25. What is the slope of the line that is tangent to

at x � 2?

a. 8

b. 12

c. 24

d. 48

26. Differentiate y = .

a.

b.

c.

d. –2x2cos(x2) + sin(x2)

27. Find when .

a.

b.

c.

d. 11x y y+( )sec( ) tan( )

dydx

=

x

y1 2 + sec ( )

dydx

=

1

1 2x y( sec ( )) +dydx

=

1x

� sec21x 2dydx

=

tan1y 2 � y � ln1x 2 � 1dy

dx

dydx

=

2x2cos1x2 2dydx

=

2x2cos1x2 2 � sin1x2 2dydx

=

xcos1x2 2 � sin1x2 2dydx

=

xsin1x2 2

y � 1x2 � 2 23

sec1x 2tan1x 2

tan21x 2

sec1x 2

f˛1x 2 � sec1x 2

�1x2 � 5x 2sin1x 2 � 12x � 5 2cos1x 2

1x2 � 5x 22

12x � 5 2cos1x 2 � 1x2 � 5 2sin1x 2

1x2 � 5 22

�sin1x 2

2x � 5

sin1x 2

2x � 5

g1x 2 �cos1x 2

x2 � 5x

xex�1dydx

=

1x � 1 2exdydx

=

xexdydx

=

exdydx

=

y � xex

5ex � 2x

5xex�1 � 2x

5xex�1 �2x

5ex �2x

f˛1x 2 � 5ex � 2ln1x 2

–POSTTEST–

158


28. Find when .

a.

b.

c.

d.

29. What is the slope of the curve

at (1,2)?

a.

b.

c.

d. 6

30. The volume of a sphere is . If the

radius increases by 3 meters per second, how

fast is the volume changing when

meters?

a.

b.

c.

d.

31. If a 10-foot ladder slides down a wall at 2 feet

per minute (see the figure that follows), how

fast does the bottom slide when the top is 6 feet

up?

a. foot per minute

b. feet per minute

c. 2 feet per minute

d. 12 feet per minute

32. Evaluate .

a. 2

b. 3

c.

d. ∞

33. Where does have a vertical asymptote?

a. y � 0

b. x � 1

c. x � e

d. no vertical asymptote

y � ex

75

limxSq

3x2 � 7x � 2x2 � 5x � 1

32

2

3

1 200, π m

sec

3

4 000, π m

sec

3

4 000

3

, πm

sec

3

2403

π msec

r � 10

V r= 4

33π

306

3

2

311

y3 � y � 3x � 3

y2 � 2xy

x2 � 2xydydx

=

y2 � 2xy

x2 � 2xydydx

=

y � 2x

x � 2ydydx

=

1y

dydx

=

x2y � xy2dy

dx

–POSTTEST–

159

ladder

10 feet

wall

y

x ground


34. Evaluate .

a. 5

b. 0

c. qd. –3x3

35. On what interval(s) is f(x) = x3 + 6x2 – 15x + 2

decreasing?

a. (4,5)

b. (�5,1)

c. (2,6) and (15,q)

d. (�q,�5) and (1,q)

36. Which of the following is the graph of

?

a.

b.

c.

d.

y � 3x � x3

limxSq

x5 � 3x3

ex � 1

–POSTTEST–

160

1

2

3

–1

–2–3 2 3

y

x

–2

–3

1–1

1

2

3

–1

–2–3 2 3

y

x

–2

–3

1–1

1

2

3

–1

–2–3 2 3

y

x

–2

–3

1–1

1

2

3

–1

–2–3 2 3

y

x

–2

–3

1–1


37. If up to 30 apple trees are planted on an acre,

each will produce 400 apples a year. For every

tree over 30 on the acre, each tree will produce

10 apples less each year. How many trees per

acre will maximize the annual yield?

a. 5 trees

b. 32 trees

c. 35 trees

d. 40 trees

38. An enclosure will be built, as depicted, with

100 feet of fencing. What dimensions will

maximize the area?

a. x = 20 ft, y = 20 ft

b. x = 25 ft, y = 10 ft

c. x = 25 ft, y = 25 ft

d. x = 50 ft, y = 10 ft

39. If and , then

what is ?

a. 6

b. 9

c. 10

d. 16

40. What is ?

a. –4

b. 0

c. 4

d. 6

�4

0

f˛1x 2 dx

�10

1

f˛1x 2 dx

�10

7

f˛1x 2 dx � 8�7

1

f˛1x 2 dx � 2

–POSTTEST–

161

y

x

1

2

3

1 2 3 4

y

x

y = f(x)

–1

–2


41. If , then what is ?

a. g ′(x) =

b. g ′(x) =

c. g ′(x) =

d. g ′(x) =

42. Evaluate .

a. 6

b. 24

c. 36

d. 40

43. Evaluate .

a.

b. �c

c.

d.

44. Evaluate

a.

b.

c.

d.

45. Evaluate .

a.

b.

c.

d.

46. Evaluate .

a.

b.

c.

d.121 ln1x 2 22 � c

1 ln1x 2 22 � c

12

x2 � c

1x

� c

� ln1x 2

x dx

3ex�1

x � 1� sin1x2 2 � c

31

22e x cx + + cos( )

3ex � cos12x 2 � c

3ex � cos12x 2 � c

� 13ex � sin12x 2 2 dx

cos1x 2 � c

�cos1x 2 � c

sin1x 2 � c

�sin1x 2 � c

�cos1x 2 dx

5x4 �2x3 � c

5x4 �2x3

16

x6 �1x

16

x6 �3x3 � c

� a x5 �1x2b dx

�3

1

16x2 � 4x 2 dx

3x2

14

x4 � c

14

t4 � c

x3

g¿ 1x 2g1x 2 � �x

0

t3 dt

–POSTTEST–

162


47. Evaluate .

a.

b.

c.

d.

48. Evaluate .

a.

b.

c.

d. 124

49. Evaluate .

a.

b.

c.

d.

50. Evaluate .

a.

b.

c.

d. xex � ex � c

xex � ex � c

12

x2ex � c

12

xex � c

�xex dx

xln1x 2 � x � c

121 ln1x 2 22 � c

1x

� c

ln11 2 � c

� ln1x 2 dx

1256

2483

2

3

�6

0

24x � 1 dx

12

x2e1x22 � c

12

xe1x22 � c

12

e1x22 � c

e1x22 � c

�xe1x22

dx

–POSTTEST–

163


Answers

1. c. Lesson 1

2. a. Lesson 1

3. d. Lesson 1

4. d. Lesson 1

5. b. Lesson 2

6. a. Lesson 2

7. b. Lesson 2

8. d. Lesson 3

9. c. Lesson 3

10. a. Lesson 3

11. c. Lesson 4

12. a. Lesson 4

13. a. Lesson 5

14. b. Lesson 5

15. c. Lesson 5

16. b. Lessons 6, 7

17. d. Lessons 6, 7

18. d. Lesson 7

19. c. Lesson 8

20. a. Lesson 8

21. a. Lesson 8

22. c. Lesson 9

23. d. Lesson 9

24. c. Lesson 9

25. d. Lesson 10

26. b. Lessons 9, 10

27. b. Lesson 11

28. c. Lesson 11

29. a. Lesson 11

30. d. Lesson 12

31. b. Lesson 12

32. b. Lesson 13

33. d. Lesson 13

34. b. Lesson 13

35. b. Lesson 14

36. d. Lesson 14

37. c. Lesson 15

38. b. Lesson 15

39. c. Lesson 16

40. c. Lesson 16

41. a. Lesson 17

42. c. Lesson 18

43. b. Lesson 18

44. b. Lesson 18

45. c. Lessons 18, 19

46. d. Lesson 19

47. b. Lesson 19

48. a. Lesson 19

49. d. Lesson 20

50. c. Lesson 20

–POSTTEST–

164


Lesson 1

1.

2.

3.

4. . Because there is no x in the

description of f, the 7 never gets used. This is

called a constant function because it is always

equal to the constant 2.

5.

6.

7. The rock is feet high after

3 seconds.

8. The profit on 100 cookies is P(100) = $39.

9.

10.

11.

=

=

12.

13.

14.

=

=− − = − + = − +2 2

22ax a

aa x a

ax a

( )( )

− + +( ) − − +( )( )x a x

a

2 25 5

− + +( ) − − +( )( )x a x

a

2 25 5

− + +( ) − − +( )( )x a x

a

2 25 5

g12x 2 � g1x 2 �

g1x2 � 2x 2 �8

x2 � 2x� 61x2 � 2x 2

−+1

2x x h( )

−+

hhx x h2 ( )

x x hhx x h

− −+

=2 ( )

x x hx x h

h

− ++

( )( )2

12

12( )x h x

h+

−=

f x h f xh

( ) ( )+ − =

f˛1x � a 2 � x2 � 2xa � a2 � 3x � 3a � 1

f˛1y 2 � y2 � 3y � 1

s13 2 � 16

h164 2 � 4

m −

= − −

= − ⋅ −

⋅ −

⋅ −

=

15

515

515

15

15

125

3

f˛17 2 � 2

h1

21

=

g1�3 2 � �20

f˛15 2 � 9

SOLUTION KEY

165

Calc2e_22_165-192_SOLKEY.qxd 11/18/11 1:05 AM Page 165

15.

16.

=

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27. (Note that t cannot equal –5.)

28.

29.

30.

31. and

32. (Note that this single interval isenough because it excludes by and –3.

Lesson 2

− 4

3

− ∞

4

3,

−( ]8 2,( , )−∞ −8

−∞ ∞( ),

−∞ ∞( ),

−∞ ∞( ),

− ∞( )5,

− ∞[ )1,

−∞ −( ) −( ) ∞( ), , , , ,3 3 5 5and

( )( )f h f x f hx

x x

o o 21

21

12

1

2

=

=−

1x3 � 2x2 � 1

� B 1x3 � 2x2 � 1

( )( ) ( ( ( )))h f g x h f g xo o = =

g h g go {( ) = −

= =

=

( ) ( ) ,16 16 16 12 1 4414

h h w w w w wo( ) = −( ) − −( )

f f z

z

zo( ) = =( )11

( )( )f h tt t

o =−1

( )( )g f xx x

o = − +1 21

3 2

( )( )f g xx x

o =− +

12 13 2

x x x xx x

3 2 326 12 8

23 6 4

+ + + − = + +

x x x x( )=

+ + +( ) −2 4 4

2

2 3

g x g x x x( ) ( ) ( )+ − = + −22

22

3 3

h x a h xa

x a xa

x a xa

aa

( ) ( ) ( )+ − =− + +( ) − − +( )

= − − + + −

= − = −

2 1 2 1

2 2 1 2 1

22

–SOLUTION KEY–

166

1

2

3

4

5

6

–1–2–3–4–5–6 1 2 3 4 5 6–1

–2

–3

–4

–5

–6

1.2.

3.4.

5.

6. 7. 8.(0,0)

(0,3)

(3,5)

( , )(–5,0)

(–3,4)

(–1,–5)

(2,–6)

y

x2

__94

__1

1

2

3

4

5

6

–2–3–4–5–6 1 2 3 4 5 6–1

y

x

7

8

9

10

12. (–2,13)

11. (0,5)

10. (1,4)

9. (3,8)

11

12

13


–SOLUTION KEY–

167

13. There is a discontinuity at x � 0. The graph of

f is decreasing on (�q,0) and on (0,q). The

graph of f is concave down on (�q,0) and

concave up on (0,q). There are no points of

inflection, no local maxima, and no local

minima. There is a vertical asymptote at x � 0

and a horizontal asymptote at y � 0.

14. There are no discontinuities. The function

increases on (�q,�3) and on (0,3), and it

decreases on (�3,0) and on (3,q). There are

local maxima at (�3,4) and (3,4), and there is

a local minimum at (0,2). The graph is concave

up on (�1,1) and concave down on (�q,�1)

and on (1,q). There are points of inflection at

(1,3) and (–1,3). There are no asymptotes.

15. There is a discontinuity at x � 1. The function

increases on (�1,1) and on (1,q), and

decreases on (�q,�1). The function is

concave up on (�q,1) and on (1,q). There is

a local minimum at (–1,�2). There are no

asymptotes, nor any points of inflection.

16. The discontinuities occur at x � �2 and x � 2.

The function increases on (0,1), (1,2) and

(2,q), and decreases on (�q,�2), (–2,–1) on

(�1,0). The point (0,2) is a local minimum.

The graph is concave up on (�2,–1), (–1,1),

and (1,2) and concave down on (�q,�2) and

(2, q). There are no points of inflection. There

are vertical asymptotes at x � �2 and x � 2,

and a horizontal asymptote at y � 0.

17. There is a discontinuity at x � �1. The

function increases on (�q,�1) and on

(�1,2), and it decreases on (2,q). There are

local maxima at (�1,3) and (2,3). The graph is

concave up on (�1,0) and concave down on

(0,q), so there is a point of inflection at (0,2).

Because the line is straight before , it

does not bow upward or downward, and thus

has no concavity. There are no asymptotes.

18. There are no discontinuities. The graph

decreases on (�q,q), is concave down on

(�q,0), and is concave up on (0, q). There is

a point of inflection at (0,0). There are

horizontal asymptotes at y � �2 and y � 2.

19. There are no discontinuities. The graph

increases on (0,2), has a local maximum at

(2,5), and decreases on (2,q). The graph is

concave down on (0,3) and concave up on

(3,q) with a point of inflection at (3,3). There

is a vertical asymptote at x � 0 and a

horizontal asymptote at y � 1.

20. The discontinuities occur at x � 2, x � 5, x � 6,

and x � 7. The function increases on (�q,1),

(4,5), and on (5,q). The function decreases on

(1,2), on (2,4), and on (6,7). There is a local

maximum at (1,2) and at (2,3). The points (4,2)

and (17,4) are a local minima. The graph is

concave up on (–q,1), (1,2), (2,5), (5,6), and

(6,7). Since the line is straight after x = 7, it does

not bow upward or downward and thus has no

concavity. There is a horizontal asymptote at y =

0 and a vertical asymptote at x = 6.

21. 3

22. 0

23.

24.w � 7

3�

7 � w�3

�94

x � �1


25.

26.

27.

28.

–SOLUTION KEY–

168

1

2

3

4

5

6

–1–2 1 2 3 4 5 6–1

y

x

y = – x + 5

(6,1)

2—3

1

2

3

–11 2 3 4 5–1

–2

–3

y

x

y = x – 2

(–1,–3)

(5,3)

1

2

3

4

56

–1–2–3–4–5–6 1 2 3 4 5 6–1

y

y = 5 (2,5) (6,5)

1

2

3

–11 2 3 4–1

–2

–3

–4

–5

–6

y

x

y = 2x – 4

(1,–2)


Lesson 3

1.

2.

3.

4.

5. 1

6.

7. 9

8. 50 = 1

9.

10.

11.

12.

13.

14.

15.

16.

17. 1

18. 2

19. 5

20. ln(1) = 0

21.

22.

23.

24. x �

25.

26.

Lesson 4

1.

2.

3.

4.

5.

6. 60°

7. 90°

8. 360°

9. 18°

10. 330°

34π

53π

3p2

p

p

6

log log log ( )

log log ( )

a a a

a a

xy

x y

x y

33

3 2 3 3

2 9 11

= −

= − = − −= + =

ln( ) ln( )

ln( ) ln( ) ln( )

ln( ) ln( ) ln( )

ln( ) ln( )ln( )

ln( )ln( )

3 3 100

3 3 100

3 5 3 100

100 5 33

1003

5

5

5

x

x

x

x

⋅ =

+ =+ =

= − = −

ln110 2

ln12 2

ln ln ln ln( )52

25

52

25

1 0

+

= ⋅

= =

ln a246b � ln14 2

ln17 # 2 2 � ln114 2

e7

e11

3

3

3

33 729

2

4 2

2

86

−

−

−

−= = =( )

44

4 2563

14

− = =

88

18

14 096

2

2 4

−= =

,

51

51

15 6256

6− = =

,

( )3 3 3 9323

3 23 2= = =⋅

12

183

=

34 � 81

6�2 �1

36

104 � 10,000

43 � 64

25 � 32

–SOLUTION KEY–

169


11. 1

12.

13. 2

14. 2

15.

16.

17.

18.

19.

20.

21.

22. �

23. �1

24. �

25.

26. �1

27. 0

28. undefined

29.

30. �

31.

32.

33. �2

34. �

35.

36.

37.

38.

39.

40. x = π

Lesson 5

1. 1

2. 1

3. 1

4. undefined

5. no

6. �1

7. 4

8. undefined

9. �1

10. yes

11. 1

12. 3

13.

14. q

q

x = 43

53

π π,

32

− 12

2

3

3

2 3

3

3

�12

32

3

2

2

2

2

2

2

2 3

3

3

3

2

2

3

2 3

3 =

3

1

3

3

3 =

3

–SOLUTION KEY–

170


15. 2

16. �2

17. 16

18. 0

19.

20.

21. 2x � 1

22.

23. �q

24. q

25.

26. �q

27.

28. �q

29. 8

30.

31. 1

32.

33. �q

34.

35. 2x

36.

37.

38. First note that

So,

39.

40. lim( )( )

( )lim ( )

( ) ( )

( ) ( )

ln( )

z

z

z

ze e

ee

e

→ →

− −− −

= − −

= − − = − −=

ln lm2

2

2 2

2

4 2

24

4 2 4

2

lim lim( )( )

( )( )

lim( ) ( )

x x

x

x x xx

x x xx x

x xx

→ →

→

+ −−

= − +− +

= ++

= ++

= =

3

3 2

2 3

3

2 159

3 53 3

53

3 3 53 3

246

4

lim( )

lim( )

lim

lim( )

lim( )

h

h

h

h

h

x h xh

x hx h x h xh

x hx h x h x

h

h x hx hh

x hx h x

→

→

→

→

→

+ −

= + + + −

= + + + −

= + +

= + + =

0

3 3

0

3 2 2 3 3

0

3 2 2 3 3

0

2 2

0

2 2 2

2 2

2 3 3 2

2 6 6 2 2

6 6 2

6 6 2 6

( ) ( ) ( ) ( )( )x h x h x h x hx h x h

x hx h x x h h x h

x hx h x h

+ = + + = + + +

= + + + + +

= + + +

3 2 2 2

3 2 2 2 2 3

3 2 2 3

2

2 2

3 3

02

0−

=

lim

lim

lim

a

a

a

a

x a x

x a x

x a x

a x a x

a

x a x x

→

→

→

+ −⋅ + +

+ +

=+ +( )

= + +( ) =

0

0

02

( )

)( )x

x

x x x→= −

− + +( ) =25

25

25 1 5

1260

lim

(

lim( )( )x

x x

x x x→

−( ) +( )− + +( )25

5 5

2 5

5 1

lim( )( )( )( )x

x xx x→

− −− +

=3

3 12 5

14

lim( )( )x

x

x x→

−− +

=2

2

2 214

72

16

e ee

− − −= =3 2 1212

2 1( )

12p6

�3p

9623

–SOLUTION KEY–

171


Lesson 6

1.

2.

3.

4.

5.

6.

.

Thus, at x � 2, the slope is .

7.

So, the slope at x = 16 is .

8.

� .

Thus, there is a slope of zero when g′(x) =

2x – 4 = 0. This happens when x � 2.

9.

� .

The slope at (2,�3) is , so

the equation of the tangent line is

.y � �41x � 2 2 � 3 � �4x � 5

h¿ 12 2 � �4

lim( )a

x a x→

− − = −0

2 2

� limaS0

�2xa � a2

a

h¿ 1x 2 � lim

aS0 1 � 1x � a 22 � 11 � x2 2

a

lim( )a

x a x→

+ − = −0

2 4 2 4

� limaS0

2xa � a2 � 4a

a

1 4 12 − − +( )x xa

g¿ 1x 2 � lim

aS0 1x � a 22 � 41x � a 2 �

a

′ =g ( )1638

′ = + −

= + − ⋅ + ++ −

= + −+ −

= // + −

=

→

→

→

→

g xx a x

a

x a xa

x a x

x a x

x a x

a x a x

a

a x a x x

a

a

a

a

( ) lim

lim

lim( )

(

lim( )

0

0

0

0

3 3

3 3 3 3

3 3

9 9

3 3

9

3 3

3

2

f ¿ 12 2 � 13

= + + = +→

lim( )a

x a x0

6 3 1 6 1

� limaS0

6xa � 3a2 � a

a

f ¿ 1x 2 � lim

aS0 31x � a 22 � 1x � a 2 � 13x2 � x 2

a

= + + =→

lim( )a

x xa a x0

2 2 23 3 3

� limaS0

13x2 � 3xa � a2 2a

a

� limaS0

x3 � 3x2a � 3xa2 � a3 � x3

a

k¿ 1x 2 � lim

aS0 1x � a 23 � x3

a

= + −+ +→

lim( )

a

x a x

x a x0

9 9

3 3

3 3

3 3

x a x

x a x

+ ++ +

= + −

→

lima

x a x

a0

3 3

′ = + −→

f xx a x

aa( ) lim

0

3 3

limaS0

0a

� 0� limaS0

10 � 10

a�

g¿ 1x 2 � lim

aS0 g1x � a 2 � g1x 2

a

� limaS0

2x � a � 2x

� limaS0

x2 � 2xa � a2 � 5 � x2 � 5

a

h¿ 1x 2 � limaS0

1x � a 22 � 5 � 1x2 � 5 2

a

� limaS0

8x � 8a � 2 � 8x � 2

a� 8

f ¿ 1x 2 � lim

aS0 81x � a 2 � 2 � 18x � 2 2

a

–SOLUTION KEY–

172

= 3

2 x

=+ +

=+→

lima x a x x x0

9

3 3

9

3 3


10.

� =

10x �2. The y-value at x � 1 is .

The slope at x � 1 is . The

equation of the tangent line is

.

Lesson 7

1.

2. = 21x 20

3.

4.

5. = 12t11

6.

7.

8.

9.

10. , so

11. , so

12. , so

13. , so

14. .

So, .

15. . So, .

16. . So, .

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32. , so ′ = + −f x x x( ) 6

152

352f x x x( ) = − −3

254

32

dy

dxx x

x x

= = 2 32 31

223

23

− −+ +

dy

du� 2u � 2u�3 � 2u �

2u3

dy

dx� �2x�2 � 2x�3 � �

2x2 �

2x3

h¿ 1u 2 � 5u4 � 16u3 � 21u2 � 4u � 8

g˛¿ 1x 2 �35

x�45 � 15x2 �

3

5x 45

� 15x2

F¿ 1x 2 � 600x99 � 500x49 � 100x24 � 20x9

′ = + +s t t et( ) 3 2 32π

dydx

x xx x

= − − − − = − + −− −4 2 3 14 6

13 43 4

( )

f ¿ 1x 2 � 24x2 � 6x

dydt

t t = −12 62

k¿ 1x 2 � �2x

g˛¿ 1t 2 �485

t3

V¿ 1r 2 � 4p r2

f ¿ 1x 2 � 30x�11 �30x11

dydx

x = 42 6

dydt

tt

= =−112

1

12

1112

1112

yt

tt t= = =−

13

14

13

14

112

′ =h t t( ) 7 6h t t( ) = 7

′ = − = −−g x x

x( )

32

3

2

52

52

g xx x x

x( ) =⋅

= = −1 112

32

32

f ¿ 1x 2 � �12

x�32 � �

1

2x˛

˛

32

f˛1x 2 �1

x˛˛

12

� x�12

dy

dx� �x�2 � �

1x2y � x�1

dy

duu

u =

1

2

1

2

12

− =y � u12

k¿ 1x 2 �14

x�34 �

1

4x˛

34

k1x 2 � x˛

14

g¿ 1x 2 � �45

x�95 � �

4

5x 95

f ¿ 1t 2 � 0

f ¿ 1x 2 � 100x99

dy

dx�

75

# x 25

dy

dt

h¿ 1x 2 � 0

g¿ 1u 2 � �5u�6 ��5u6

dy

dx

f ¿ 1x 2 � 5x4

y � 121x � 1 2 � 7 � 12x � 5

k¿ 11 2 � 12

k˛11 2 � 7

lima

x a→

+ +( )0

10 5 2 limaS0

10xa � 5a2 � 2a

a

limaS0

51x � a 22 � 21x � a 2 � 15x2 � 2x 2

a�

k¿ 1x 2 �

–SOLUTION KEY–

173


33. , ,

,

and

34.

35.

36. , , and

Lesson 8

1. pay rate in dollars per hour

2. fuel economy in miles per gallon

3. baby’s growth rate in pounds per month

4. rate at which the radius shrinks in inches per

hour

5. , so . So, it is increasing at

the rate of 1 foot per year.

6. , so . So, it is

decreasing at the rate of 6 feet per day.

7. , so

. So, the profit is increasing at

the rate of $3,750 per car sold.

8. , so the cost would

increase at a rate of $2.13 per inch at the

instant when x = 3 inches.

9. After 3 seconds, it is at meters from

the start. Note that ν(t) = 3t 2 + 4t + 10 and

a(t) = 6t + 4. At that moment, it is traveling at

meters per second and accelerating

at meters per second per second.

10. The position function is ,

the velocity function is , and the

acceleration is a constant . Since

s(2) = 0, it will hit the ground when t � 2

seconds and be traveling at feet

per second (downward) at that instant.

11. The position function is

and the velocity function is .

The bullet will stop in the air when the velocity

is zero. This happens at t � 25 seconds, when

the bullet is feet in the air.

12. The position function is

, so after 4 seconds, it is feet

off the ground. It has therefore fallen 296 feet

by this moment. Since ν(t) = –32t – 10, it is

traveling feet per second

(downward) at this moment.

13.

14.

15.

16.

17. because is a constant

18. Because

, the point is . Since

f ′(x) = cos(x) – sin(x), the slope is

, so the equation is

.� �x �p

2� 1� a x �

p

2b � 1

y �f ¿ ap2b � �1

ap

2,1b1 � 0 � 1

sin ap

2b � cos a

p

2b �f a

p

2b �

cos15 2h¿ 1x 2 � �sin1x 2

r¿ 1u 2 �12

cos1u 2 �12

sin1u 2

′ = − − +

= − − +

− −g x x x x

x xx

( ) sin( )

sin( )

653

6 5

3

343

3 43

π

π

f ¿ 1t 2 � 3cos1t 2 �2t2

dy

dx� 20x4 � 10sin1x 2

v14 2 � �138

s14 2 � 7041,000

s1t 2 � �16t2 � 10t �

s125 2 � 10,000

v1t 2 � �32t � 800

s1t 2 � �16t2 � 800t

v12 2 � �64

a1t 2 � �32

v1t 2 � �32t

s1t 2 � �16t2 � 64

a13 2 � 22

v13 2 � 49

s13 2 � 75

′ = − ≈C ( ) . .3 4 883

2 13

′ =P ( ) ,50 3 750

′ = − +P x x x( ) ,3

10120 9 0002

′ = −L ( )7 6′ = − +L t t( ) 2 8

′ =h ( )5 1′ =h tt

( )25

2

d3y

dt3 �209

t�83

d2y

dt2 � �43

t�53

dy

dt� 2t�2

3

d y

dxx x x

3

36 5 4180 48 6= − + −− − −

s– 1t 2 � �32

f 1x 2 � 24x�5

f ‡ 1x 2 � �6x�4

f – 1x 2 � 2x�3f ¿ 1x 2 � �x�2

–SOLUTION KEY–

174


19.

20.

21.

22. . (Note that e–3 is a

constant.)

23. . (Note that ln(π) is a

constant.)

24. , so

25.

26.

Lesson 9

1. = x(2cos(x) – x

sin(x))

2.

3.

4.

= x(6ln(x) + 3 – 20x2)

5.

6.

7.

8.

9.

10.

11.

12.

�

13.So, ,

so the slope is 1.

14.

15.

16.

17.

18.

19.

14et � 1 2 1t3 � 2t � 1 2 � 13t2 � 2 2 14et � t 2

1t3 � 2t � 1 22

dy

dt�

x x

x

4

2

5 1ln( )

ln( )

−( )( )

dy

dx�

5x4ln1x 2 � 1x# x5

1 ln1x 2 22�

5x4ln1x 2 � x4

1 ln1x 2 22

f ¿ 1x 2 �11 � 1

x 2 1ex � 1 2 � ex1x � ln1x 2 2

1ex � 1 22

2x1x2 � 1 2 � 2x1x2 � 1 2

1x2 � 1 22�

4x1x2 � 1 22

f ¿ 1x 2 �

16x � 5 2 1x3 � 10x � 7 2

13x2 � 5x � 2 22

13x2 � 10 2 13x2 � 5x � 2 2

13x2 � 5x � 2 22�h¿ 1x 2 �

y � �p1x � p 2 � �px � p2

f ¿ 10 2 � 210 2e0 � e010 22 � 1 � 1

′ = + +f x xe e xx x( ) .2 12

3x3cos1x 2 � 3x4sin1x 2ln1x 2

12x3ln1x 2cos1x 2 �

a1x

cos1x 2 � sin1x 2ln1x 2b # 3x4

g¿ 1x 2 � 12x3ln1x 2cos1x 2 �

dy

dx� exsin1x 2 � 1exsin1x 2 � cos1x 2ex 2 # x

� 2cos1x 2sin1x 2

cos1x 2sin1x 2 � cos1x 2sin1x 2f ¿ 1x 2 �

dy

dxx x

xx

x x

5 1 ln( ) + 1

= − ⋅ ⋅

= − −

1

15 1

2

2 ln( )

′

= −( ) + +( ) +( )h t

t t t t t t

( )

sin( ) cos( ) cos( ) sin( )3 42 2

dy

dx�

8sin1x 2

x� 8ln1x 2cos1x 2 � sin1x 2

′ = ⋅ − ⋅

− ⋅ − ⋅

−

− −

k x x x x x

x x x x

( ) cos( ) sin( )

cos( ) sin( )

12

14

12

12

14

54

′ = + + +

= + +

h u e u u u e

e u u

u u

u

( ) ( ) ( )

( )

2

2

3 2 3

5 3

g˛¿ 1x 2 � 6xln1x 2 �1x13x2 2 � 20x3

dy

dxx x = −π(cos ( ) sin ( ))2 2

dy

dt� 24t˛

2et � et # 8t˛

3 � 8t˛

2et13 � t 2

f ¿ 1x 2 � 2xcos1x 2 � sin1x 2 # x2

f ¿ 110 2 �1

10

g˛

11002 1x 2 � 3ex

f – 1x 2 � e˛

x �1x2f ¿ 1x 2 � ex �

1x

k¿ 1u 2 �152

x 32 � 5ex

h¿ 1x 2 �1

22x�

8x

dy

dx� �sin1x 2 � 10e˛

x � 8

g˛¿ 1t 2 �12t

� 2t

f ¿ 1x 2 � 1 � 2x � 3x2 � ex

–SOLUTION KEY–

175


20. g ′t

21.

22.

23.

24.

25.

26.

27.

28. , so

.

29.

30.

31.

32.

33.

34.

Lesson 10

1.

2.

3.

4.

5.

�2x � 9

22x2 � 9x � 1

121x2 � 9x � 1 2�

12 # 12x � 9 2g¿ 1x 2 �

dy

du�

721u5 � 3u4 � 7 2

52 # 15u4 � 12u3 2

18t7 � 27t2 � 3 2

101t8 � 9t3 � 3t � 2 2 #h¿ 1t 2 �

dy

dx� 31x2 � 8x � 9 22 # 12x � 8 2

f ¿ 1x 2 � 418x3 � 7 23 # 124x2 2

′ =

+

− ⋅ +

− −

j x

x x x x x x x

x

( )

sec( ) sec( ) tan( )

sec ( )

13

14

23

34

13

14

2

′ = ⋅ + ⋅

+

⋅ ⋅

h t e tt

e t t

e t

t t

t

( ) ln( ) tan( ) sec ( )

ln( )

1 2

′ = +

= +( )g x e x x x e

e x x

x x

x

( ) sec( ) sec( ) tan( )

sec( ) tan( )1

f ¿ 1x 2 � tan1x 2 � x # sec21x 2

�csc21x 2��1

sin21x 2�

��sin21x 2 � cos21x 2

sin21x 2

ddxa

cos1x 2

sin1x 2b

ddx1cot1x 2 2 �

= − = − ⋅

= −

cos( )

sin ( ) sin( )cos( )sin( )

csc( )cot( )

x

x xxx

x x

2

1

ddxa

1sin1x 2

bddx1csc1x 2 2 �

′ = + = + ==

f e e e e e e e( ) ln( )2 2 31

{

′ = ⋅ +f x x x x( ) ln( )2

dydx

e e x e e x e

d y

dxe x e e x

x x x

x x x

= ⋅ + ⋅ + = ⋅ + +

= ⋅ + + ⋅ = ⋅ +

1 1

1 1 22

2

( )

( ) ( )

f˛¿ 1x 2 �12xex � ex # x2 2cos1x 2 � sin1x 2 # x2ex

cos21x 2

�ln1x 2 � 1 � xln1x 2

ex

1 ln1x 2 � 1 2ex � xex˛ln1x 2

ex # ex

dy

dx�

′ =

+

⋅ −

⋅ + ⋅

⋅ +( )=

h t

tt

t t t t t t

t

t t

( )

sin ( )

cos( ) sin( ) cos( ) sin( ) ln( )

sin ( )

sin( )cos( )

11 2

2

4

6 744444 844444

dydx

x x x x

x x=

− ⋅ −( ) − ⋅ +( )( )+( ) −( )

02 2

cos( ) cos( ) sin( ) sin( )

sin( ) cos( )

π π

π π

′ = ⋅ − − − ⋅

−( )g u

u u u u u

u u( )

cos( ) ( ) ( ) sin( )3 2

32

3 3 2

3

dydx

x x x x

x

x

x

=− +

− −

+

= −

+

32

132

1

1

3

1

12

32

12

32

32

12

32

2

2

= −

= −

3

3

2 3

2 2

2 3

2

π ππ

π

t t t t

t

t t t t

t

sin( ) cos( )

sin ( )

sin( ) cos( )

sin ( )

–SOLUTION KEY–

176


6.

7.

8.

9.

10. h′(x) =

11.

12.

13. , so

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

Lesson 11

1. , so

2. , so

3.

4.

5.

6.

7.dy

dx�

�sin1x 2

sec21y 2� �sin1x 2cos21y 2

dy

dxe ee e

x x

y y = +

+22

2

2

dy

dx�

12x3 � 12y � 8

dy

dx�

1x

1 � 121y

�22y

2x2y � x

dy

dx�

4cos1y 2

� 4sec1y 2

dy

dx�

cos1x 2

3y2 � 13y2 # dy

dx�

dy

dx� cos1x 2

dy

dx�

4x3 � 831y � 1 22

31y � 1 22 # dy

dx� 4x3 � 8

′ = −−

⋅ − −

−

⋅⋅ ⋅ −( ) + ⋅ −

−( )

h xe

e

e

e

e x e e e

e

x

x

x

x

x x x x

x

( ) cos sin21

1

1

1

8 1 1

1

4 4

4 4

2

2 2

2 2

�3u

sec1 ln18u3 2 2tan1 ln18u3 2 2

sec1 ln18u3 2 2tan1 ln18u3 2 2 # 18u3

# 24u2k¿ 1u 2 �

dy

dx� cos1sin1sin1x 2 2 2 # cos1sin1x 2 2 # cos1x 2

g¿ 1t 2 �sec21et � 1 2 # et

tan1et � 1 2

dy

dx� 41e9x2�2x�1 23 # 1e9x2�2x�1 2 # 118x � 2 2

� �24cos218x 2sin18x 2

1�sin18x 2 2 # 8f ¿ 1x 2 � 3cos218x 2 #

dydx

x x= ( ) ⋅ −( )sec cos( ) sin( )2

′ = +( ) ⋅ −( )s u u u u u( ) sin( ) cos( ) cos( ) sin( )32

dydx x x

x x x= ⋅ + ⋅( )1sin( )

sin( ) cos( )

dy

dxe x exx e = ⋅ + −2 2 12 2

sec110x2 � ex 2tan110x2 � ex 2 # 120x � ex 2

f ¿ 1x 2 �

dy

dx� e2x � 2xe2x

= −2 2 22

θ θ θθ

cos( ) sin( )

cos12u 2 # 2 # u � 1 # sin12u 2

u2f ¿ 1u 2 �

g˛¿ 1x 2 �121ex � e�x 2g1x 2 �

121ex � e�x 2

f ¿ 1x 2 � ex � 2e2x � 3e3x

dy

dx�

51 ln1x 2 24

x

π πcos( )x

dy

dt�

33t � 5

′ = ( ) ⋅−

g x x x( ) tan( ) sec ( )12

12 2

′ = ( ) ⋅f x xx

( ) sec2 1

2

dy

dx�

131ex � 1 2�

23 # ex �

ex

31ex � 1 223

–SOLUTION KEY–

177


8. , so

9.

10.

11.

12.

13. , so

14.

15. , so at

(�3,1), the tangent slope is .

16. , so at (1,�2) the tangent slope

is .

17. , so at (4,2) the tangent slope is

1.

18. , so at

the tangent slope is .

19. at , so the tangent

equation is .

20. , so at (0,0) the tangent slope is

. As such, the equation of the tangent line is

y = x.

Lesson 12

1.

2.

3.

4.

5.

6. 2A # dAdt

� 2B # dBdt

� 2C # dCdt

dzdt

�45

x # dxdt

�45

y # dy

dt�

35x2

# dxdt

1y

# dy

dt� ex # dx

dt� 2x # dx

dt# y2 � 2y # dy

dt# x2

1

22x# dx

dt�

1

22y# dy

dt� 30x2 # dx

dt� 7 # dx

dt

4y3dy

dt� 6x

dxdt

� �sin1y 2dy

dt

dy

dt� 51x3 � x � 1 24 # a 3x2dx

dt�

dxdtb

− 13

− 13

dydx

ee ee

e e

x

x y

x

x y

=−

+

++ 1

y x = 3

2 3 1

2 6−

+ π

12 6

,π

dy

dx =

3

2 3

− 4 2π

22 4

,π

dydx

y y

= −

⋅

1

2 2 2 2

π πcos( ) sin sin( )

dydx

yy

= −

−−

13

3 52

− 49

dydx

xy

= − −−

3 13 3

2

2

dy

dx� 1

3y2 # dy

dx� 2x � 2y # dy

dx� 5 # dy

dx

dy

dx�

3x2cos1y 2

sec1y 2tan1y 2 � 9 � x3sin1y 2

dy

dx�

1 � y �1y

�xy

2 � x � 1�

y2 � y3 � y

�x � xy2 � y2

y � x # dydx

y2 � y �dy

dx# x � 1 �

dy

dx

dy

dx

x y xy

x y x = −

−4 2

4

3 4

2 3 4

dy

dx�

521y � x2 23

� 2x

dydx x y y

= −− −

1

3 2 22( )

dydx

x x yy x y

= − + −− + −

cos( ) cos( )cos( ) cos( )

dy

dx

x y

x yx y

=

1

2 +

1 1

2 +

1

2 + −=

− 1

dy

dx x y

dy

dx =

1

2 + 1 +

–SOLUTION KEY–

178

Though tempting, don’t cancel the terms cos(x – y).


7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17. Because , A is increasing at the rate

of 172 square feet per minute when I � 20.

18. , so R is increasing at the rate of

per hour at this instant.

19. , so the area is decreasing at the

rate of 65 square feet per minute.

20. , so the area shrinks by

320p square inches per hour.

21. , so the radius grows at

feet per hour.

22. , so each side is shrinking at the rate of

4 inches per minute.

23. , so each side is growing at the rate of

inch per second.

24. , so the base increases at the rate of

7 inches per hour.

25. If the height is y and the base is x, then

and . After 6

seconds, y � 6 and , so x � 8.

Because , , so

. The end of the board is moving at

the rate of of a foot each hour along the

ground.

26. If the base is x and the hypotenuse (length of

the string) is s, then . Using

this, when s � 260, x must be 240. Because

, we can calculate that .

Thus, the string is being let out at 12 feet

per second.

Lesson 13

1. 0

2.

3.

4.q

52

45

dsdt

� 12dxdt

� 13

x2 � 1002 � s2

34

dxdt

� �34

218 2dxdt

� 216 2 11 2 � 0dy

dt� 1

x2 � 62 � 100

2xdxdt

� 2ydy

dt� 0x2 � y2 � 102

dbdt

� 7

1240

dedt

= − 1240

dsdt

� 4

52p

� 0.796drdt

�5

2p

dAdt

in

hour= − 320

2

π

dAdt

= − 65

2764

dRdt

= 2764

dAdt

� 172

dBdt

� 28

dKdt

� 24

dy

dt� �

43

dxdt

� �3512

dDdt x y

xdxdt

ydydt

xdxdt

ydydt

x y

=+

⋅ ⋅ + ⋅

=⋅ + ⋅

+

1

22 2

2 2

2 2

dSdt

ededt

= ⋅12

= ⋅ + ⋅

12

dbdt

hdhdt

b

dAdt

�12

# dbdt

# h �dhdt

# 12

b

dCdt

� 2p # drdt

dAdt

� 8p r # drdt

dVdt

� 4p r2 # drdt

–SOLUTION KEY–

179


5. 0

6.

7. �q

8. q

9. 1

10. 0

11. 3

12. q

13. �q

14. q

15. 0

16. 0

17. q

18. 0

19. 0

20.

21. vertical asymptote at x � 4, horizontal

asymptote at y � 1, sign diagram:

22. vertical asymptotes at x � 2 and x � �2,

horizontal asymptote at y � 0, sign diagram:

23. vertical asymptote at x � �3, horizontal

asymptote at y � 1, sign diagram:

24. vertical asymptotes at x � 1 and x � 3, hori-

zontal asymptote at y � 0, sign diagram:

25. vertical asymptote at x = –5, horizontal asymp-

tote at y = 0, sign diagram:

26. vertical asymptotes at x = –2 and x = –1, no

horizontal asymptote, sign diagram:

27. q

28. q

29. q

30. �q

31. �q

32. �q

–1–2 0

m(x)+- +-

–5

j(x) + -

�47

�89

–SOLUTION KEY–

180

++ -f (x)

–2 4

++ --g(x)

–2 2 3

+++ -h(x)

–3 –1 1

++ --k(x)

1 31__2

–


Lesson 14

1. has no asymptotes. ,

thus there is a local minimum at (15,�215).

Because , the graph is always

concave up.

2. has no asymptotes.

, so there

is a local maximum at (�2,4). Because

, the graph is always concave

down.

g– 1x 2 � �2

g¿ 1x 2 � �4 � 2x � �212 � x 2

g1x 2

f – 1x 2 � 2

f˛¿ 1x 2 � 2x � 30f˛1x 2

–SOLUTION KEY–

181

1

2

3

–1

–2–3 1 2 3–1

y

x

–2

g(x) = –4x – x2


concavity

4(–2,4)

(0,0)

–2

y

x

f(x) = x2 – 30x + 10

15increasingdecreasing

concavity

5 10 15 20

25

–25

–50

–75

–100

–125

–150

–175

–200

–225

–250

(0,10)

(15,–215)



,

so there is a local maximum at (�2,49) and a

local minimum at (3,�76). Because

, there is a point of

inflection at .


, so there

is a local minimum at (�1,�2) and a local

maximum at (1,2). , so there is a

point of inflection at (0,0).

k– 1x 2 � �6x

k¿ 1x 2 � 3 � 3x2 � 311 � x 2 11 � x 2

k1x 2

1

2

27

2,−

12 a x �12b12x � 6 �

h– 1x 2 �

h¿ 1x 2 � 6x2 � 6x � 36 � 61x � 3 2 1x � 2 2

h1x 2

–SOLUTION KEY–

182

1

2

3

–1

–2–3

1

2 3

–1

y

x

–2

–3

k(x) = 3x – x3

0


concavity

(–1,–2)

(1,2)

(0,0)–1 125

50

75

–25

–2–3 1 2 3–1

y

x

–50

–75

h(x) = 2x3 – 3x2 – 36x + 5


concavity

–4

(–2,49)

,

(3, –76)

–2 3

1—2

1—2–27—2



, so there

is a local minimum at (6,�427).

, so there

are points of inflection at (0,�5) and at

(4,�251).

6. has a vertical asymptote at x � �2 and a

horizontal asymptote at y � 1. The first

derivative is , and the

second is .g– 1x 2 ��4

1x � 2 23

g¿ 1x 2 �2

1x � 2 22

g1x 2

f – 1x 2 � 12x2 � 48x � 12x1x � 4 2

f ¿ 1x 2 � 4x3 � 24x2 � 4x21x � 6 2

f˛1x 2

–SOLUTION KEY–

183

–3 1 2 3

y

x

f(x) =x4 – 8x3 + 5


concavity

4 5 6 7 8 9–2 –1

50

–50

–100

–150

–200

–250

–300

–350

–400

–450

(0,5)

(4,–251)

(6,–427)

6

0 4

______

1

2

3

–1–2–3 1 2 3–1

y

x

–2

–3

g(x) = xx + 2


concavity

–4

4

4 5–4–5

–2

–2

(–3,3)

(–1, –1)


7. has vertical

asymptotes at x � �3 and x � 3, and a

horizontal asymptote

at y � 0. Because

, there is a local maximum at

. The second derivative is

.

8. has vertical

asymptotes at x � 1 and x � �1, and a

horizontal asymptote at y � 0.

The first derivative is

,

and the second derivative is

. There is a point of

inflection at (0,0).

k– 1x 2 �2x1x2 � 3 2

1x � 1 231x � 1 23

k¿ 1x 2 � �x2 � 11x2 � 1 22

� �x2 � 1

1x � 1 221x � 1 22

k1x 2 �x

1x � 1 2 1x � 1 2

6x2 � 181x � 3 231x � 3 23

h– 1x 2 �6x2 � 181x2 � 9 23

�

a 0,�19b

�2x1x � 3 221x � 3 22

h¿ 1x 2 ��2x1x2 � 9 22

�

h1x 2 �1

1x � 3 2 1x � 3 2

–SOLUTION KEY–

184

–2–3 1 2 3–1

y

x

k(x) = –——— xx2 – 1

4 5–4–5

1

2

–1

–2

0

–1 1

–1 1

(2, )2—3

(–2, ) 2 – — 3–2

–3 1

2

3–1

y

x

h(x) = 1______x2 – 9

4 5–4–5

1

2

–1

–2

–3 30

–3 3

0, 1—9–


9. has a vertical asymptote at x � 0 but no

horizontal asymptote. Because

, there is a

local maximum at (�1,�2) and a local

minimum at (1,2). The second derivative is

.

10. has a horizontal asymptote at y � 0 but

no vertical asymptotes. Because

, there is

a local minimum at and a local

maximum at . Because

, there are points

of inflection at , (0,0), and

.33

4,

33

4,

= − ++

2 3 3

12 3

x x x

x

( )( )

( )

f – 1x 2 �2x1x2 � 3 2

1x2 � 1 23

a ˇˇˇˇ1,12b

a ˇˇˇˇ�1,�12b

f˛¿ 1x 2 �1 � x2

1x2 � 1 22�11 � x 2 11 � x 2

1x2 � 1 22

f˛1x 2

j– 1x 2 �2x3

j¿ 1x 2 �x2 � 1

x2 �1x � 1 2 1x � 1 2

x2

j1x 2

–SOLUTION KEY–

185

(– , –

–2–3 1 2 3–1

y

x

–

f(x) = x2+1

______x

0

1–1

–

1,

–1,–

3

3

3 ___4 )

( ), 3___4

33

1—2

1—2

1—2

1—2

–2–3 1 2 3–1

y

x

j(x) = x______x2+ 1

4 5–4–5

1

2

–1

–2

0

3

4

–3

–4

(1,2)

(–1,–2)

–1 10


Lesson 15

1. when x � 10.

Using the first derivative test, is

positive, so the function increases to x � 10

and , so the

function decreases afterward, thus x � 10

maximizes the profit.

2. If x is the number of trees beyond 30 that are

planted on the acre, then the number of

oranges produced will be:

O(x) = (number of trees) (yield per tree)

� (30 + x)(500 � 10x) � 15,000 + 200x � 10x2.

The derivative is zero

when x � 10. Using the second derivative test,

is negative, so this is maximal.

Thus, x � 10 more than 30 trees should be

planted, for a total of 40 trees per acre.

3. The total sales will be figured as follows:

Sales = (number of copies)(price per copy),

so S(x) = (20 + x)(100 – x) = 2,000 + 80x – x2

where x is the number of copies beyond 20.

The derivative S′(x) = 80 – 2x is zero when x

� 40. Because S″(x) = –2, this is maximal by

the second derivative test. Thus, the artist

should make x � 40 more than 20 paintings,

for a total of 60 paintings in order to maximize

sales.

4. After x days, there will be pounds of

watermelon, which will valued at cents

per pound. Thus, the price after x days will be

P(x) = (200 + 5x)(90 – x) = 18,000 + 250x – 5x2

cents. The derivative is P(x) = 250 – 10x, which

is zero when x � 25. Because P(x) is clearly

positive when x is less than 25 and negative

afterward, this is maximal by the first

derivative test. Thus, the watermelons will

fetch the highest price in 25 days.

5. The area is and the total fencing is

. Thus, , so the

area function can be written as follows:

A(y) = xy = (200 – 26) ⋅ y = 200y – 2y 2. The

derivative A′(y) = 200 – 4y is zero when y � 50.

Because the second derivative is A″(y) = –4, this

is an absolute maximum. Thus, the optimal

dimensions for the pen are y � 50 feet and

feet.

6. Here, and the total fencing is

. Because , the

area function can be written as follows:

A(y) = (150 – 5y)y = 150y – 5y2. The

derivative A′(y) = 150 – 10y is zero when y �

15. Because A″(y) = –10, this is an absolute

maximum. Thus, the optimal dimensions are y

� 15 feet and therefore

feet.

7. Because Volume = πr2h = 16π, it follows

that . Thus, the surface area function is

A(r) =


when , so . Thus, the only

point of slope zero is when r � 2. The second

derivative is , which is

positive when r � 2. Thus, by the second

derivative test, r � 2 is the absolute minimum.

Thus, a radius of r � 2 inches and a height of

inches will minimize the surface

area.

h �16r2 � 4

′′ = +A ( )rr

464

3π π

r3 � 8432

2π π

rr

=

′ −V x x( ) = 150 3

22

2 216

2322

22π π π π

r rr

rr

+

= +

h �16r2

x � 150 � 5115 2 � 75

x � 150 � 5y5y � x � 150

Area � xy

x � 200 � 2y � 200 � 2150 2 � 100

x � 200 � 2y4y � 2x � 400

Area � xy

90 � x

200 � 5x

′′ = −O ( )x 20

′ = −O ( )x x200 20

P¿ 1100 2 � �1

10�

1100

� �9

100

P¿ 11 2 � 9,000

P¿ 1x 2 � �1,000

x2 �10,000

x3 � 0

–SOLUTION KEY–

186


8. Because the box has a square bottom, let x be

both its length and its width, and let y be the

height. Thus, the volume is

and the surface area is

(the top, the four sides, and the bottom). And

because , the height

. Thus, the volume

is

.


when . Negative lengths are

impossible, therefore this is zero only when

x � 10. By the second derivative test,

is negative when x � 10, so

this is a maximum. The corresponding height

is feet,

so the largest box is a cube with all sides of

length 10 feet.

9. Because the box has a square bottom, let x be

both the length and width, and y be the height.

The area of each side is thus xy, so the cost to

build four of them at ten cents a square foot is

0.10(4xy) = 0.4xy dollars. The area of the top is

, so it will cost dollars to build.

Similarly, it will cost dollars to build the

base. The total cost of the box is therefore

Cost = 0.4xy + 8x2. Because the volume is

, we know . Thus, the

cost function can be written:

.

The derivative:

is zero only when or x � 10.

By the second derivative test,

is positive when

x � 10, so this is the absolute minimum. The

cheapest box will be built when x � 10 feet and

y � 400 feet.

′′ =

+C 6( ),

xx

32 0001

3

x3 � 1,000

′ = −

+C 6( ),

xx

x16 000

12

�16,000

x� 8x2

C . ( ),

x xx

x=

+0 440 000

82

2

y �40,000

x2x2y � 40,000

7x2

x2x # x � x2

y �15010

�102

� 10

′′ = −V ( )x x3

x2 � 100

′ −V x x( ) = 150 3

22

V ( )x xx

xx x= −

= −2 31502

15012

y �600 � 2x2

4x�

150x

�x2

Area � 2x2 � 4xy � 600

Area � x2 � 4xy � x2

Volume � x2y

–SOLUTION KEY–

187


10. Inside the margins, the area is:

.

The total area of the page is , so

. Therefore, the area is

.


when , thus when x � 12 ignore

negative lengths. The second derivative

is negative when x � 12,

so this is the absolute maximum. Thus, the

dimensions that maximize the printed area are

x � 12 inches and y � 8 inches.

Lesson 16

1. 2

2.

3.

4. 2p

5. 3

6. 3 � 2p

7.

8. 0

9.

10. 4

11. 0

12.

13. 16

14. 6

15. 0

16.

17. 35

18. 48

19. 5

20.21. 7

22. 7

23.24.25. 8

26.27. 25

28. 11

29. –1

30. 1

Lesson 17

1. 0 (Note that , for any constant a.)

2.

3. 8

4.

5. 28

6.

7. 0

8. 7

9. 14

852

332

52

g x dxa

a( ) =∫ 0

�17

�1

�11

�3

212

�12

�32

�32

72

32

′′ −A = 576

3( )x

x

x2 � 144

′ − +A = 288

2( )x

x2

� 102 � 2x �288

x

A = ( )x xx

xx

962 3

966

− −

+

y �96x

xy � 96

� 1x � 3 2 1y � 2 2 � xy � 2x � 3y � 6

Area = x y−

−232

2( )

–SOLUTION KEY–

188


10. 21

11. 28

12. 35

13. 10

14.

15. 36

16. 20

17.

18.

19.

20.

21.

22.

23.

24. 1

Lesson 18

1.

2.

3.

4.

5. 0

6.

7.

8.

9.

10.

11.

12. 40

13.

14.

15.

16.

17. 48

18. 11

19.

20.

21.

22. 58

23. 208

24.

25.

26. 3ex �12

x4 � c

13

x3 � 5sin1x 2 � c

913

x133 �

5611

x117 � c

− + − +− −12

43

12

4 3 2t t t c

�6

14

t˛

12 � 3t˛

3 �1˛

2t˛

2 � c

2x3 � 5x2 � 5x � c

34

23

43

32x x c− +

83

u3 � c

95

x5 � c

5x � c

34

u 43 � c

45

x 54 � c

38

x˛

˛

83 � c

−[ ] = −

=−

−

−

−

−

tt

1

4

1

4

11 3

4

1�2

t�2 � c � �1

2t2 � c

13

723

0

6

x

=

17

u7 � c

113

x13 � c

15

5x c+

− 22

12

22

2,0003

383

163

23

�4

–SOLUTION KEY–

189


27.

28.

29.

30.

31.

32.

Lesson 19

1. , by substituting u = x 5 + 1

2. , by substituting u = 4x + 3

3. , by substituting u = x 3 + 1

4.

5. , by substituting u = x 2 – 1

6. , by substituting u = 2x + 1

7. , by substituting u = 1 – x

8. ln 3x3 – 5x, by substituting u = 3x3 – 5x

9. , by substituting u = x 4

10. , by substituting

u = 3x 4 – 2x + 1


u = 4x 2 + 5x – 1


u = 4x 2 + 5


u = 4x + 10

14. Using , the solution is

. Using , the solution

is . Because

, these solutions will

be the same if the second is greater

than the first one.

15. , by substituting u = sin(x)

16. , by substituting u = 4x

17.


u = 7x – 2

19. cos(ex) + c, by substituting u = ex (the one

inside sin(ex))

20. , by substituting u = ln(x)


22. , by substituting (the one

occurring in the term )

23. , by substituting u = cos(x)

24. , by substituting u = 1 + e2x12

1 2ln( )+ +e cx

− +lncos( )x c

e x

u x=2e cx +

ln ln( )x c +

141 ln1x 2 24 � c

�17

cos17x � 2 2 � c

4sin1x 2 � c

14

sin14x 2 � c

13

sin31x 2 � c

12

�c

sin21x 2 � cos21x 2 � 1

�12

cos21x 2 � c

u � cos1x 212

sin21x 2 � c

u � sin1x 2

12

4 10ln x c + +

�1

1614x2 � 5 22� c

1414x2 � 5x � 1 24 � c

23x4 � 2x � 1 � c

12

sin1x4 2 � c

56

13

2 132( )x c+ +

38

1243( )x c− +

14

x4 �92

x2 � 4x � c

115

14414x � 3 211 � c

1401x5 � 1 28 � c

8 06

56sin( )x[ ] =π

π

4 4 4 4 3 4 2 42

33 2e e ex[ ] = − = − =

ln( )

ln( )ln( ) ln( ) ( ) ( )

a12

� e1b � 10 � e0 2 � e �12

− + +5 2cos( )x e cx

12u2 � 2cos1u 2 � c

2 2 2

4 2 4 2 2

22ln ln ln

ln( ) ln( )

u e e

e ee

e[ ] = −

= − = − =

–SOLUTION KEY–

190


Lesson 20

1. , done by parts with

2. , by parts with

3. , by the substitution

4. , by parts with


6. , using

parts twice, first with u = x2, and then with

u = 2x

7. , by basic integration


9. , by substituting u = e3x + 9

10. ,

by parts with

11. , evaluating

by parts with

12. , substituting

13. , by parts

with

14. , by parts with

15. , by substituting u = cos(x)


17. , by basic integration

18. e–cos(x) + c, by substituting u = –cos(x)



21. First, let u = six(x).

Then, .

From an earlier example,

. So, we get

sin(x) ⋅ ln(sin(x)) – sin(x) + c.

22. , by parts twice,

plus the trick from the last example in the

lesson.

121e

xsin1x 2 � e xcos1x 2 2 � c

ln( ) ln( )u du u u u c= − +�

ln( )u du�cos( )ln sin( )x x dx( ) =�

u � cos1x 2�231cos1x 2 2

32 � c

u �1x

�e 1x � c

13

ln x c +

−( )

+1

44

ln( )xc

− +14

4cos ( )x c

u � x�xe�x � e�x � c

u � x

23

x˛1x � 1 232 �

4151x � 1 2

52 � c

u � x � 1231x � 1 2

32 � c

u � ln1x 2ln( )x dx�ln ln( )x x x x c + − +

u � ln1x 2

a14

x4 �32

x2 � xb ln1x 2 �1

16x4 �

34

x2 � x � c

13

93ln( )e cx + +

u � x3 � 113

ex3�1 � c

13

x3 � cos1x 2 � c

�x2cos1x 2 � 2xsin1x 2 � 2cos1x 2 � c

u � ln1x 2121 ln1x 2 22 � c

u � x � 3

1x � 3 2sin1x 2 � cos1x 2 � c

u � x2�12

cos1x2 2 � c

u � x�xcos1x 2 � sin1x 2 � c

u � ln1x 2

16

x6ln1x 2 �1

36x6 � c

–SOLUTION KEY–

191


acceleration the rate at which the velocity of a

moving object is increasing or decreasing

additive rule parts of a function added together

can be differentiated separately:

antiderivative given a function , the

anti-derivative is a function such that

.

asymptote a line that a graph flattens out

toward. It occurs as either an infinite limit or a

limit as .

Chain Rule

closed interval the set of all the real numbers

between and including two endpoints, like all x

such that

composition the process of plugging one func-

tion into another. The composition of functions

f and g is .

concave down when the graph of a function

bows downward, like a frown

concave up when the graph of a function bows

upward, like a smile

concavity the way a graph curves either upward

or downward

Constant Coefficient Rule a constant c multi-

plied in front of a function is unaffected by dif-

ferentiation:

Constant Rule the derivative of a constant is

zero.

continuous function one whose graph can be

drawn without picking up the pencil

cosecant abbreviated csc; see trigonometry

cosine abbreviated cos; see trigonometry

cotangent abbreviated cot; see trigonometry

critical points points of slope zero, points where

the derivative is undefined, and endpoints of the

domain

decreasing when the graph of function goes

down from left to right

definite integral the area between a graph

and the x-axis from x � a to x � b

where area below the x-axis counts as negative,

written

degrees measure the size of angles in such a way

that a complete circle is 360°

�a

b

f˛1x 2dx

y � f˛1x 2

ddx1c # f˛1x 2 2 � c # f ¿ 1x 2

( )( ) ( ( ))f g x f g xo =

a � x � b

ddx1f˛1g1x 2 2 2 � f ¿ 1g1x 2 2 # g¿ 1x 2

x → ±∞

g¿ 1x 2 � f˛1x 2

g1x 2

f˛1x 2

ddx1f1x 2 � g1x 2 2 � f ¿ 1x 2 � g¿ 1x 2

GLOSSARY

193

Calc2e_23_193-196_GLOS 11/18/11 1:07 AM Page 193

derivative the derivative of function is

,

which is the slope of the tangent line at point

.

discontinuity a break in a graph

domain the set of all the real numbers at which a

function can be evaluated

e a transcendental number approximately equal to

2.71828 . . .

explicit a function is explicit if its formula is

known exactly.

exponent an exponent says how many times a

factor is multiplied by itself. In the case of roots,

the exponent is a fraction.

First Derivative Test if a function increases to a

point and then decreases afterward, then that

point is a local maximum. If the function

decreases to a point and then increases afterward,

then the point is a local minimum.

function a mathematical object that assigns one

number in its range to every number in its

domain

Fundamental Theorem of Calculus if

, then . Thus,

where .

graph a visual depiction of a function where the

height of each point is the value assigned to the

number on the horizontal axis

horizontal asymptote a horizontal line toward

which the graph flattens out as or

implicit a function is implicit if it was defined in

an indirect manner so that its exact formula is

unknown.

implicit differentiation the process of taking a

derivative of both sides of an equation and using

the Chain Rule with , ,

, and so on

increasing when the graph of a function goes up

from left to right

indefinite integral represents the antiderivative:

if and only if

integral see either definite integral or indefinite

integral

L’Hôpital’s Rule If and

, then .

The same is true when .

limit the limit means that the values

of get very close to L as x gets close to a.

limit from the left means that the

values of are close to L when x is close to,

and less than, a.

limit from the right means that the

values of are close to L when x is close to,

and greater than, a.

limits at infinity means that

the values of get close to as x gets

really big. If large negative values of x are used and

gets close to , then .

limits of integration the limits of the integral

are a and b.

local maximum the highest point on a graph in

that immediate area, like a hilltop

�b

a

f˛1x 2dx

lim ( )x

f x L→−∞

= y � Ly � f˛1x 2

y � Ly � f˛1x 2

lim ( )x

f x L→∞

=

f˛1x 2

limxSa�

f˛1x 2 � L

f˛1x 2

limxSa�

f˛1x 2 � L

f˛1x 2

limxSa

f˛1x 2 � L

limx→−∞

lim( )

( )

( )

( )x x

f x

g x

f x

g x→∞ →∞= ′

′ limlim ( )

xg x

→∞= ± ∞

lim ( )x

f x→∞

= ± ∞

g¿ 1x 2 � f˛1x 2� f˛1x 2dx � g1x 2 � c

ddt1x 2 �

dxdt

ddx1y 2 �

dy

dxddx1x 2 � 1

x → −∞x → ∞

g¿ 1x 2 � f˛1x 2�b

a

f˛1x 2dx � g1b 2 � g1a 2

g¿ 1x 2 � f˛1x 2g1x 2 � �x

0

f˛1t 2dt

1x,f˛1x 2 2

dy

dx� f ¿ 1x 2 � lim

aS0

f˛1x � a 2 � f˛1x 2

a

y � f˛1x 2

–GLOSSARY–

194


local minimum the lowest point on a graph in that

immediate area, like a valley

natural logarithm the inverse of the expo-

nential function . Thus, if and only if

.

oscillate to repeatedly go back and forth across a

range of values

point of inflection a point on a graph where the

concavity changes

point-slope formula the equation of a straight

line through with slope m is

.

polynomial the sum of powers of a variable,

complete with constant coefficients. For example,

and are

both polynomials.

position function gives the mark on a line where a

moving object is at a given time

Power Rule

Product Rule

Pythagorean theorem the squares of the legs of a

right triangle add up to the square of the

hypotenuse.

Quotient Rule

radians measure the size of angles in such a way

that a complete circle is 2p radians

range the set of all the numbers that can be the

value of a function

rate of change how fast a quantity is increasing or

decreasing

secant abbreviated sec; see trigonometry

rational function a rational function is the quo-

tient of two polynomials. For example,

is a rational function.

second derivative the derivative of the first derivative

Second Derivative Test a point of slope zero is the

maximum if the second derivative is negative at

the point, and a minimum if the second derivative

is positive at the point.

sign diagram tells where an expression is positive

and negative

sine abbreviated sin; see trigonometry

slope a measure of steepness of a straight line. It is

the amount the y-value goes up or down with

each step to the right.

slope-intercept formula the equation of a straight

line with slope m that crosses the y-axis at y � b is

.

Squeeze Theorem if and

, then .

substitution an integration technique used to

reverse the Chain Rule

tangent abbreviated tan; see trigonometry

tangent line a straight line that indicates the direc-

tion of a curve at a given point

third derivative the derivative of the second

derivative

trigonometric identities ,

, ,

, and sin2 1x 2 � cos2 1x 2 � 1cot1x 2 �cos1x 2

sin1x 2

csc1x 2 �1

sin1x 2sec1x 2 �

1cos1x 2

tan1x 2 �sin1x 2

cos1x 2

limxSa

g1x 2 � LlimxSa

f˛1x 2 � L � limxSa

h1x 2

f˛1x 2 � g1x 2 � h1x 2

y � mx � b

8x3 � 10x � 45x � 2

ddxa

f˛1x 2

g1x 2b �

f ¿ 1x 2 # g1x 2 � g¿ 1x 2 # f˛1x 2

1g1x 2 22

ddx1f˛1x 2 # g1x 2 2 � f ¿ 1x 2 # g1x 2 � g¿ 1x 2 # f˛1x 2

ddx1xn 2 � n # xn�1

10x7 � 12x5 � 4x2 � xx2 � 3x � 5

y � m1x � x1 2 � y1

1x1,y1 2

ey � x

y � ln1x 2ex

ln1x 2

–GLOSSARY–

195


trigonometry the study of functions formed by

dividing one side of a right triangle by another.

When the right triangle has angle x, the

hypotenuse has length H, the side adjacent to x

has length A, and the side opposite x has length O,

then

The derivatives are:

unit circle the circle of radius 1 centered at the

origin

velocity the rate of change of a moving object at a

particular time

vertical asymptote a vertical line x = a that a graph

more closely resembles as the inputs x approach a

ddx1cot1x 2 2 � �csc2 1x 2

ddx1tan1x 2 2 � sec2 1x 2

ddx1csc1x 2 2 � �csc1x 2cot1x 2

ddx1sec1x 2 2 � sec1x 2tan1x 2



cot1x 2 �AO

tan1x 2 �OA

csc1x 2 �HO

sec1x 2 �HA

cos1x 2 �AH

sin1x 2 �OH

–GLOSSARY–

196


calculus success in 20 minutes a day2ndedition[1]

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