chapter 4 the exponential and natural logarithm functions

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 55

Chapter 4

The Exponential and Natural Logarithm

Functions


Exponential Functions

The Exponential Function ex

Differentiation of Exponential Functions

The Natural Logarithm Function

The Derivative ln x

Properties of the Natural Logarithm Function

Chapter Outline


§ 4.1




Properties of Exponential Functions

Simplifying Exponential Expressions

Graphs of Exponential Functions

Solving Exponential Equations

Section Outline


Exponential Function

Definition Example

Exponential Function: A function whose exponent is the independent variable

xy 3


Properties of Exponential Functions


Simplifying Exponential Expressions

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Write each function in the form 2kx or 3kx, for a suitable constant k.

(a) We notice that 81 is divisible by 3. And through investigation we recognize that 81 = 34. Therefore, we get

x

xx

ba

22

2

81

1

152

.3333

1

81

1 224242

4

2xxx

xx

(b) We first simplify the denominator and then combine the numerator via the base of the exponents, 2. Therefore, we get

.222

2

22

2 61151

1515xxx

x

x

x

x


Graphs of Exponential Functions

Notice that, no matter what b is (except 1), the graph of y = bx has a y-intercept of 1. Also, if 0 < b < 1, the function is decreasing. If b > 1, then the function is increasing.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve the following equation for x. 054532 xxx

054532 xxx This is the given equation.

04325 xx Factor.

0365 xx Simplify.

03605 xx Since 5x and 6 – 3x are being multiplied, set each factor equal to zero.

xx 205 5x ≠ 0.


§ 4.2

The Exponential Function e x


e

The Derivatives of 2x, bx, and ex

Section Outline


The Number e

Definition Example

e: An irrational number, approximately equal to 2.718281828, such that the function f (x) = bx has a slope of 1, at x = 0, when b = e

xexf


The Derivative of 2x



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Calculate.

2

2x

x

dx

d

173.025.0693.02 2 2

2

mdx

d

x

x


The Derivatives of bx and ex



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the equation of the tangent line to the curve at (0, 1).x

x

ex

ey

We must first find the derivative function and then find the value of the derivative at (0, 1). Then we can use the point-slope form of a line to find the desired tangent line equation.

x

x

ex

ey

This is the given function.

x

x

ex

e

dx

dy

dx

dDifferentiate.

2x

xxxx

ex

exdxd

eedxd

ex

dx

dy

Use the quotient rule.



2

1x

xxxx

ex

eeeex

dx

dy

Simplify.

CONTINUECONTINUEDD

2

1x

xxx

ex

eexe

dx

dy

Factor.

2

1x

x

ex

xe

dx

dy

Simplify the numerator.

Now we evaluate the derivative at x = 0.

11

1

10

11

0

101220

0

0

20

e

e

ex

xe

dx

dy

xx

x

x



CONTINUECONTINUEDDNow we know a point on the tangent line, (0, 1), and the slope of that line, -1. We will now use the point-slope form of a line to determine the equation of the desired tangent line.

11 xxmyy This is the point-slope form of a line.

011 xy (x1, y1) = (0, 1) and m = -1.

1 xy Simplify.


§ 4.3

Differentiation of Exponential Functions


Chain Rule for eg(x)

Working With Differential Equations

Solving Differential Equations at Initial Values

Functions of the form ekx

Section Outline



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Differentiate. 32 2xexg x

This is the given function. 32 2xexg x

Use the chain rule. xedx

dxexg xx 223 222

Remove parentheses.

x

dx

de

dx

dxexg xx 223 222

Use the chain rule for exponential functions.

2223 222 xx exexg


Working With Differential Equations

Generally speaking, a differential equation is an equation that contains a derivative.


Solving Differential Equations

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Determine all solutions of the differential equation

.3

1yy

The equation has the form y΄ = ky with k = 1/3. Therefore, any

solution of the equation has the form

yy3

1

xCey 3

1

where C is a constant.


Solving Differential Equations at Initial Values

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Determine all functions y = f (x) such that y΄ = 3y and f (0) = ½.

The equation has the form y΄ = ky with k = 3. Therefore,yy 3

xCexf 3

for some constant C. We also require that f (0) = ½. That is,

.02

1 003 CCeCef

So C = ½ and

.2

1 3xexf


Functions of the form ekx


§ 4.4

The Natural Logarithm Function


The Natural Logarithm of x

Properties of the Natural Logarithm

Exponential Expressions


Solving Logarithmic Equations

Other Exponential and Logarithmic Functions

Common Logarithms

Max’s and Min’s of Exponential Equations

Section Outline


The Natural Logarithm of x

Definition Example

Natural logarithm of x: Given the graph of y = ex, the reflection of that graph about the line y = x, denoted y = ln x



1) The point (1, 0) is on the graph of y = ln x [because (0, 1) is on the graph of y = ex].

2) ln x is defined only for positive values of x.

3) ln x is negative for x between 0 and 1.

4) ln x is positive for x greater than 1.

5) ln x is an increasing function and concave down.


Exponential Expressions

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Simplify.xe ln23ln

Using properties of the exponential function, we have

.3332ln2ln2ln2

3lnln23ln

xeeeee

ee

xxxx



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve the equation for x. 4322 xx ee

This is the given equation. 4322 xx ee

Remove the parentheses.4322 xx ee

Combine the exponential expressions.

4322 xxe

Add.42 xe

Take the logarithm of both sides. 4lnln 2 xe

Simplify.4ln2 x

Finish solving for x.4ln2 x


Solving Logarithmic Equations

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve the equation for x.82ln5 x

This is the given equation.82ln5 x

Divide both sides by 5.6.12ln x

Rewrite in exponential form.

6.12 ex

Divide both sides by 2.477.22

6.1

e

x


Other Exponential and Logarithmic Functions


Common Logarithms

Definition Example

Common logarithm: Logarithms to the base 10

2100log10

31000log10

4000,10log10


Max’s & Min’s of Exponential Equations

EXAMPLEEXAMPLE

The graph of is shown in the figure below. Find the coordinates of the maximum and minimum points.

xexxf 211



xexxf 211 This is the given function.

CONTINUECONTINUEDDAt the maximum and minimum points, the graph will have a slope of zero. Therefore, we must determine for what values of x the first derivative is zero.

xx edx

dxx

dx

dexf 22 11 Differentiate using the product

rule.

xx exxexf 2112 Finish differentiating.

121 xxexf x Factor.

1210 xxex Set the derivative equal to 0.

012010 xxex Set each factor equal to 0.

110 xxex Simplify.



CONTINUECONTINUEDDTherefore, the slope of the function is 0 when x = 1 or x = -1. By looking at the graph, we can see that the relative maximum will occur when x = -1 and that the relative minimum will occur when x = 1.

Now we need only determine the corresponding y-coordinates.

1011111 12 eef

472.0

41

211111

212

eeef

Therefore, the relative maximum is at (-1, 0.472) and the relative minimum is at (1, -1).


§ 4.5

The Derivative of ln x


Derivatives for Natural Logarithms

Differentiating Logarithmic Expressions

Section Outline


Derivative Rules for Natural Logarithms



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Differentiate. 22 1ln xe

This is the given expression. 22 1ln xe

Differentiate. 22 1ln xedx

d

Use the power rule. 1ln1ln2 22 xx edx

de

Differentiate ln[g(x)]. 11

11ln2 2

22

x

xx e

dx

d

ee

Finish. xx

x ee

e 22

2 21

11ln2



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

The function has a relative extreme point for x > 0. Find the coordinates of the point. Is it a relative maximum point?

xxxf /1ln

This is the given function. xxxf /1ln

Use the quotient rule to differentiate.

2

11ln1

x

xx

xxf

Simplify. 2

ln

x

xxf

Set the derivative equal to 0.2

ln0

x

x



Set the numerator equal to 0.

xln0

CONTINUECONTINUEDDThe derivative will equal 0 when the numerator equals 0 and the denominator does not equal 0.

Write in exponential form.10 ex

To determine whether the function has a relative maximum at x = 1, let’s use the second derivative.

This is the first derivative. 2

ln

x

xxf

Differentiate.

22

2 2ln1

x

xxx

xxf



CONTINUECONTINUEDD

Simplify. 4

ln2

x

xxxxf

Factor and cancel. 3

ln21

x

xxf

Evaluate the second derivative at x = 1.

11

021

1

1ln211

3

f

Since the value of the second derivative is negative at x = 1, the function is concave down at x = 1. Therefore, the function does indeed have a relative maximum at x = 1. To find the y-coordinate of this point

.11/101/11ln1 f

So, the relative maximum occurs at (1, 1).


§ 4.6




Simplifying Logarithmic Expressions


Logarithmic Differentiation

Section Outline


Simplifying Logarithmic Expressions

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Write as a single logarithm.zyx ln3ln2

1ln5

This is the given expression.zyx ln3ln2

1ln5

Use LIV (this must be done first).

3215 lnlnln zyx

Use LIII.3

21

5

lnln zy

x

Use LI.

321

5

ln zy

x

Simplify.

21

35

lny

zx



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Differentiate.

14

21ln

32

x

xxx

This is the given expression.

14

21ln

32

x

xxx

Rewrite using LIII. 14ln21ln 32 xxxx

Rewrite using LI. 14ln2ln1lnln 32 xxxx

Rewrite using LIV. 14ln2ln31ln2ln2

1 xxxx

Differentiate.

14ln2ln31ln2ln2

1xxxx

dx

d



Distribute.

CONTINUECONTINUEDD

14ln2ln31ln2ln2

1

xdx

dx

dx

dx

dx

dx

dx

d

Finish differentiating.414

1

2

13

1

12

1

2

1

xxxx

Simplify.14

4

2

3

1

2

2

1

xxxx



Definition Example

Logarithmic Differentiation: Given a function y = f (x), take the natural logarithm of both sides of the equation, use logarithmic rules to break up the right side of the equation into any number of factors, differentiate each factor, and finally solving for the desired derivative.

Example will follow.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Use logarithmic differentiation to differentiate the function.

5

43

4

32

x

xxxf

This is the given function. 5

43

4

32

x

xxxf

Take the natural logarithm of both sides of the equation.

5

43

4

32lnln

x

xxxf

Use LIII. 543 4ln32lnln xxxxf

Use LI. 543 4ln3ln2lnln xxxxf



Use LIV. 4ln53ln42ln3ln xxxxf

CONTINUECONTINUEDD

Differentiate. 4

5

3

4

2

3ln

xxxxf

xfxf

dx

d

Solve for f ΄(x).

4

5

3

4

2

3

xxxxfxf

Substitute for f (x).

4

5

3

4

2

3

4

325

43

xxxx

xxxf

chapter 4 the exponential and natural logarithm functions

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