chapter 11 - differentiation

29
INTRODUCTORY MATHEMATICAL ANALYSIS INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences 2007 Pearson Education Asia Chapter 11 Chapter 11 Differentiation Differentiation

Upload: muhammad-bilal-khairuddin

Post on 16-Apr-2017

685 views

Category:

Business


2 download

TRANSCRIPT

Page 1: Chapter 11 - Differentiation

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences

2007 Pearson Education Asia

Chapter 11 Chapter 11 DifferentiationDifferentiation

Page 2: Chapter 11 - Differentiation

2007 Pearson Education Asia

INTRODUCTORY MATHEMATICAL ANALYSIS

0. Review of Algebra

1. Applications and More Algebra

2. Functions and Graphs

3. Lines, Parabolas, and Systems

4. Exponential and Logarithmic Functions

5. Mathematics of Finance

6. Matrix Algebra

7. Linear Programming

8. Introduction to Probability and Statistics

Page 3: Chapter 11 - Differentiation

2007 Pearson Education Asia

9. Additional Topics in Probability10. Limits and Continuity

11. Differentiation12. Additional Differentiation Topics13. Curve Sketching14. Integration15. Methods and Applications of Integration16. Continuous Random Variables17. Multivariable Calculus

INTRODUCTORY MATHEMATICAL ANALYSIS

Page 4: Chapter 11 - Differentiation

2007 Pearson Education Asia

• To compute derivatives by using the limit definition.

• To develop basic differentiation rules.

• To interpret the derivative as an instantaneous rate of change.

• To apply the product and quotient rules.

• To apply the chain rule.

Chapter 11: Differentiation

Chapter ObjectivesChapter Objectives

Page 5: Chapter 11 - Differentiation

2007 Pearson Education Asia

The Derivative

Rules for Differentiation

The Derivative as a Rate of Change

The Product Rule and the Quotient Rule

The Chain Rule and the Power Rule

11.1)

11.2)

11.3)

Chapter 11: Differentiation

Chapter OutlineChapter Outline

11.4)

11.5)

Page 6: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation

11.1 The Derivative11.1 The Derivative• Tangent line at a point:

• The slope of a curve at P is the slope of the tangent line at P.

• The slope of the tangent line at (a, f(a)) is

hafhaf

azafzfm

haz

0tan limlim

Page 7: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.1 The Derivative

Example 1 – Finding the Slope of a Tangent Line

Find the slope of the tangent line to the curve y = f(x) = x2 at the point (1, 1).Solution: Slope = 211lim11lim

22

00

h

hh

fhfhh

• The derivative of a function f is the function denoted f’ and defined by

h

xfhxfxz

xfzfxfhxz

0

limlim'

Page 8: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.1 The Derivative

Example 3 – Finding an Equation of a Tangent Line

If f (x) = 2x2 + 2x + 3, find an equation of the tangent line to the graph of f at (1, 7).Solution:

Slope

Equation

24322322limlim'22

00

x

hxxhxhx

hxfhxfxf

hh

16

167

xy

xy

62141' f

Page 9: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.1 The Derivative

Example 5 – A Function with a Vertical Tangent Line

Example 7 – Continuity and Differentiability

Find . Solution:

xdxd

xh

xhxxdxd

h 21lim

0

a. For f(x) = x2, it must be continuous for all x.b. For f(p) =(1/2)p, it is not continuous at p = 0, thus the derivative does not exist at p = 0.

Page 10: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation

11.2 Rules for Differentiation11.2 Rules for Differentiation• Rules for Differentiation:

RULE 1 Derivative of a Constant:

RULE 2 Derivative of xn:

RULE 3 Constant Factor Rule:

RULE 4 Sum or Difference Rule

0cdxd

1 nn nxxdxd

xcfxcfdxd '

xgxfxgxfdxd ''

Page 11: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.2 Rules for Differentiation

Example 1 – Derivatives of Constant Functionsa.

b. If , then .

c. If , then .

03 dxd

5xg

4.807623,938,1ts

0' xg

0dtds

Page 12: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.2 Rules for Differentiation

Example 3 – Rewriting Functions in the Form xn

Differentiate the following functions:Solution:a.

b.

xy

xx

dxdy

21

21 12/1

xx

xh 1

2/512/32/3

23

23' xxx

dxdxh

Page 13: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.2 Rules for Differentiation

Example 5 – Differentiating Sums and Differences of Functions

Differentiate the following functions: xxxF 53 a.

x

xxx

xdxdx

dxdxF

2115

2153

3'

42/14

2/15

3/1

4 54

b.z

zzf

3/433/43

3/1

4

35

3154

41

54

'

zzzz

zdzdz

dzdzf

Page 14: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.2 Rules for DifferentiationExample 5 – Differentiating Sums and Differences of Functions

8726 c. 23 xxxy

7418

)8()(7)(2)(6

2

23

xxdxdx

dxdx

dxdx

dxd

dxdy

Page 15: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.2 Rules for Differentiation

Example 7 – Finding an Equation of a Tangent LineFind an equation of the tangent line to the curve when x = 1.

Solution: The slope equation is

When x = 1,

The equation is

xxy 23 2

2

12

23

2323

xdxdy

xxxx

xy

5123 2

1

xdxdy

45

151

xy

xy

Page 16: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation

11.3 The Derivative as a Rate of Change11.3 The Derivative as a Rate of Change

Example 1 – Finding Average Velocity and Velocity

• Average velocity is given by

• Velocity at time t is given by

t

tfttftsvave

t

tfttfvt

0lim

Suppose the position function of an object moving along a number line is given by s = f(t) = 3t2 + 5, where t is in seconds and s is in meters.

a.Find the average velocity over the interval [10, 10.1].b. Find the velocity when t = 10.

Page 17: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.3 The Derivative as a Rate of ChangeExample 1 – Finding Average Velocity and Velocity

Solution:a. When t = 10,

b. Velocity at time t is given by

When t = 10, the velocity is

m/s 3.601.0

30503.3111.0

101.101.0

101.010

fffft

tfttftsvave

tdtdsv 6

m/s6010610

tdt

ds

Page 18: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.3 The Derivative as a Rate of Change

Example 3 – Finding a Rate of Change

• If y = f(x),

then

xxxx

xfxxfxy

to from interval the over x to respect with y of change of rate average

xrespect toy with

of change of rate ousinstantanelim

0 xy

dxdy

x

Find the rate of change of y = x4 with respect to x, and evaluate it when x = 2 and when x = −1. Solution: The rate of change is .

34xdxdy

Page 19: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.3 The Derivative as a Rate of Change

Example 5 – Rate of Change of VolumeA spherical balloon is being filled with air. Find the rate of change of the volume of air in the balloon with respect to its radius. Evaluate this rate of change when the radius is 2 ft.Solution: Rate of change of V with respect to r is

When r = 2 ft,

22 4334 rr

drdV

ftft1624

32

2

rdr

dV

Page 20: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.3 The Derivative as a Rate of Change

Applications of Rate of Change to Economics

• Total-cost function is c = f(q).

• Marginal cost is defined as .

• Total-revenue function is r = f(q).

• Marginal revenue is defined as .

dqdc

dqdr

Relative and Percentage Rates of Change

• The relative rate of change of f(x) is .

• The percentage rate of change of f (x) is

xfxf '

%100'xfxf

Page 21: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.3 The Derivative as a Rate of Change

Example 7 – Marginal Cost

If a manufacturer’s average-cost equation is

find the marginal-cost function. What is the marginal cost when 50 units are produced?Solution: The cost is

Marginal cost when q = 50,

qqqc 5000502.00001.0 2

5000502.00001.0

5000502.00001.0

23

2

qqq

qqqqcqc

504.00003.0 2 qqdqdc

75.355004.0500003.0 2

50

qdq

dc

Page 22: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.3 The Derivative as a Rate of Change

Example 9 – Relative and Percentage Rates of Change

11.4 The Product Rule and the Quotient Rule11.4 The Product Rule and the Quotient Rule

Determine the relative and percentage rates of change of

when x = 5.Solution:

2553 2 xxxfy

56' xxf

%3.33333.0

7525

55' change %

255565'

ff

f

The Product Rule xgxfxgxfxgxf

dxd ''

Page 23: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.4 The Product and Quotient Rule

Example 1 – Applying the Product Rule

Example 3 – Differentiating a Product of Three Factors

Find F’(x).

153412435432

543543'

543

22

22

2

xxxxxx

xdxdxxxxx

dxdxF

xxxxF

Find y’.

26183

432432'

)4)(3)(2(

2

xx

xdxdxxxxx

dxdy

xxxy

Page 24: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.4 The Product and Quotient Rule

Example 5 – Applying the Quotient RuleIf , find F’(x).

Solution:

The Quotient Rule

2

''xg

xgxfxfxgxgxf

dxd

1234 2

x

xxF

22

2

2

22

1232122

12234812

12

12343412'

xxx

xxxx

x

xdxdxx

dxdx

xF

Page 25: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.4 The Product and Quotient Rule

Example 7 – Differentiating Quotients without Using the Quotient Rule

Differentiate the following functions.

5

6352'

52 a.

22

3

xxxf

xxf

44

33

7123

74'

74

74 b.

xxxf

xx

xf

455

41'

3541

435 c.

2

xf

xx

xxxf

Page 26: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.4 The Product and Quotient Rule

Example 9 – Finding Marginal Propensities to Consume and to Save

If the consumption function is given by determine the marginal propensity to consume and the marginal propensity to save when I = 100.

Solution:

Consumption Function

dIdC consume to propensity Marginal

consume to propensity Marginal - 1 save to propensity Marginal

10

325 3

I

IC

2

32/1

2

32/3

1013233105

10

103232105

IIII

I

IdIdII

dIdI

dIdC

536.01210012975

100

IdIdC

Page 27: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation

11.5 The Chain Rule and the Power Rule11.5 The Chain Rule and the Power Rule

Example 1 – Using the Chain Rulea. If y = 2u2 − 3u − 2 and u = x2 + 4, find dy/dx.

Solution:

Chain Rule:

Power Rule:

dxdu

dudy

dxdy

dxdunuu

dxd nn 1

xu

xdxduu

dud

dxdu

dudy

dxdy

234

4232 22

xxxxxxdxdy 26821342344 322

Page 28: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.5 The Chain Rule and the Power RuleExample 1 – Using the Chain Rule

Example 3 – Using the Power Rule

b. If y = √w and w = 7 − t3, find dy/dt.

Solution:

3

22

32/1

72

323

7

tt

wt

tdtdw

dwd

dtdy

If y = (x3 − 1)7, find y’.Solution:

622263

3173

121317

117'

xxxx

xdxdxy

Page 29: Chapter 11 - Differentiation

2007 Pearson Education Asia

Chapter 11: Differentiation11.5 The Chain Rule and the Power Rule

Example 5 – Using the Power Rule

Example 7 – Differentiating a Product of Powers

If , find dy/dx.

Solution:

21

2

xy

22

2112

2

2221

xxx

dxdx

dxdy

If , find y’.

Solution:

452 534 xxy

2425215342

4531053412

453534'

2342

424352

524452

xxxx

xxxxx

xdxdxx

dxdxy