Download Exponential Function

Chapter 4
The Exponential and Natural
Logarithm Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 1 of 55
Chapter Outline

Exponential Functions

The Exponential Function ex

Differentiation of Exponential Functions

The Natural Logarithm Function

The Derivative ln x

Properties of the Natural Logarithm Function
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 2 of 55
§ 4.1
Exponential Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 3 of 55
Section Outline

Exponential Functions

Properties of Exponential Functions

Simplifying Exponential Expressions

Graphs of Exponential Functions

Solving Exponential Equations
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 4 of 55
Exponential Function
Definition
Example
Exponential Function:
A function whose
exponent is the
independent variable
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y   3
x
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 5 of 55
Properties of Exponential Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 6 of 55
Simplifying Exponential Expressions
EXAMPLE
Write each function in the form 2kx or 3kx, for a suitable constant k.
a   1 
 81 
x 2
25 x 1
b 
2  2 x
SOLUTION
(a) We notice that 81 is divisible by 3. And through investigation we recognize
that 81 = 34. Therefore, we get
1
 
 81 
x 2
1
 4 
3 
x2
 
 3 4
x 2
 3  4  x 2   3  2 x.
(b) We first simplify the denominator and then combine the numerator via the
base of the exponents, 2. Therefore, we get
25 x 1 25 x 1
 1 x  25 x 11 x   26 x.
x
22
2
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 7 of 55
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.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 8 of 55
Solving Exponential Equations
EXAMPLE
Solve the following equation for x.
2  3x 5x  4  5x  0
SOLUTION
2  3x 5x  4  5x  0
5 x 2  3x   4  0
5 x 6  3x  0
5x  0
5x  0
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6  3x  0
2x
This is the given equation.
Factor.
Simplify.
Since 5x and 6 – 3x are being
multiplied, set each factor
equal to zero.
5x ≠ 0.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 9 of 55
§ 4.2
The Exponential Function e x
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 10 of 55
Section Outline

e

The Derivatives of 2x, bx, and ex
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 11 of 55
The Number e
Definition
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
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Example
f x   e x
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 12 of 55
The Derivative of 2x
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 13 of 55
Solving Exponential Equations
EXAMPLE
Calculate.
 
d x
2
dx
x  2
SOLUTION
 
d x
2
dx
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 m  22  0.6930.25  0.173
x  2
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 14 of 55
The Derivatives of bx and ex
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 15 of 55
Solving Exponential Equations
EXAMPLE
ex
Find the equation of the tangent line to the curve y 
x  ex
at (0, 1).
SOLUTION
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.
ex
y
x  ex
This is the given function.
d
d  ex 

 y   
x 
dx
dx  x  e 
Differentiate.
d
d


xe  e e
xe 
dy
dx
dx

dx
x  e 
x
x
x
x 2
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x
Use the quotient rule.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 16 of 55
Solving Exponential Equations
CONTINUED




dy x  e x e x  e x 1  e x

2
dx
x  ex


 
 
dy e x x  e x  1  e x

2
dx
x  ex

Simplify.

Factor.
dy e x x  1

dx x  e x 2


Simplify the numerator.
Now we evaluate the derivative at x = 0.
dy
dx
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
x 0
e x x  1
x  e 

x 2
x 0
e 0 0  1
0  e 
0 2

1 1  1

 1
2
1
0  1
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 17 of 55
Solving Exponential Equations
CONTINUED
Now 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.
y  y1  mx  x1 
y 1  1x  0
y  x 1
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This is the point-slope form of
a line.
(x1, y1) = (0, 1) and m = -1.
Simplify.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 18 of 55
§ 4.3
Differentiation of Exponential Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 19 of 55
Section Outline

Chain Rule for eg(x)

Working With Differential Equations

Solving Differential Equations at Initial Values

Functions of the form ekx
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 20 of 55
Chain Rule for eg(x)
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 21 of 55
Chain Rule for eg(x)
EXAMPLE
Differentiate.

g x   e 2 x  2x

3
SOLUTION

g x   e 2 x  2x

3
This is the given function.






g x   3 e  2 x  2 x 
2

d 2 x
e  2x
dx

2  d
d

g x   3 e  2 x  2 x   e  2 x  2 x 
dx 
 dx

g x   3 e 2 x  2 x  2e 2 x  2
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2
Use the chain rule.
Remove parentheses.
Use the chain rule for
exponential functions.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 22 of 55
Working With Differential Equations
Generally speaking, a differential equation
is an equation that contains a derivative.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 23 of 55
Solving Differential Equations
EXAMPLE
1
3
Determine all solutions of the differential equation y  y.
SOLUTION
1
3
The equation y   y has the form y΄ = ky with k = 1/3. Therefore, any
solution of the equation has the form
y  Ce
1
x
3
where C is a constant.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 24 of 55
Solving Differential Equations at Initial Values
EXAMPLE
Determine all functions y = f (x) such that y΄ = 3y and f (0) = ½.
SOLUTION
The equation y  3 y has the form y΄ = ky with k = 3. Therefore,
f x   Ce3 x
for some constant C. We also require that f (0) = ½. That is,
1
 f 0  Ce30  Ce0  C.
2
So C = ½ and
f x  
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1 3x
e .
2
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 25 of 55
Functions of the form ekx
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 26 of 55
§ 4.4
The Natural Logarithm Function
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 27 of 55
Section Outline

The Natural Logarithm of x

Properties of the Natural Logarithm

Exponential Expressions

Solving Exponential Equations

Solving Logarithmic Equations

Other Exponential and Logarithmic Functions

Common Logarithms

Max’s and Min’s of Exponential Equations
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 28 of 55
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
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 29 of 55
Properties of the Natural Logarithm
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 30 of 55
Properties of the Natural Logarithm
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.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 31 of 55
Exponential Expressions
EXAMPLE
Simplify.
e ln 3 2 ln x
SOLUTION
Using properties of the exponential function, we have
e
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ln 3 2 ln x
eln3
3
3
3
 2 ln x  2ln x  2 ln x  2 .
e
e
e e
e x
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 32 of 55
Solving Exponential Equations
EXAMPLE
Solve the equation e x   e 23 x  4 for x.
2
SOLUTION
e   e
x 2
2 3 x
4
e 2 x  e 2 3 x  4
e 2 x   2 3 x   4
e 2x  4
Remove the parentheses.
Combine the exponential
expressions.
Add.
 
Take the logarithm of both sides.
2  x  ln 4
Simplify.
ln e 2x  ln 4
x  2  ln 4
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This is the given equation.
Finish solving for x.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 33 of 55
Solving Logarithmic Equations
EXAMPLE
Solve the equation 5 ln 2x  8 for x.
SOLUTION
5 ln 2x  8
ln 2x  1.6
2 x  e1.6
e1.6
x
 2.477
2
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This is the given equation.
Divide both sides by 5.
Rewrite in exponential form.
Divide both sides by 2.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 34 of 55
Other Exponential and Logarithmic Functions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 35 of 55
Common Logarithms
Definition
Example
Common logarithm:
Logarithms to the
base 10
log 10 100  2
log 10 1000  3
log 10 10,000  4
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 36 of 55
Max’s & Min’s of Exponential Equations
EXAMPLE
The graph of f x   1  x  12 e x is shown in the figure below. Find the
coordinates of the maximum and minimum points.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 37 of 55
Max’s & Min’s of Exponential Equations
CONTINUED
At 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.
f x   1  x  1 e x
2
f x   e x 
This is the given function.
d
x  12 x  12  d e x
dx
dx
Differentiate using the product
rule.
f x   e x  2x  1  x  1 e x
Finish differentiating.
f x   e x x  12  x  1
Factor.
2
0  e x x  12  x  1
e x  0 x 1  0 2  x 1  0
e x  0 x  1 x  1
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Set the derivative equal to 0.
Set each factor equal to 0.
Simplify.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 38 of 55
Max’s & Min’s of Exponential Equations
CONTINUED
Therefore, 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.
f 1  1  1  1 e1  1  0  e  1
2
f  1  1   1  1 e
2
1
2

 2
 1 
e
 1 
4
 0.472
e
Therefore, the relative maximum is at (-1, 0.472) and the relative minimum is at
(1, -1).
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 39 of 55
§ 4.5
The Derivative of ln x
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 40 of 55
Section Outline

Derivatives for Natural Logarithms

Differentiating Logarithmic Expressions
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 41 of 55
Derivative Rules for Natural Logarithms
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 42 of 55
Differentiating Logarithmic Expressions
EXAMPLE
Differentiate.
ln e
2x

1
2
SOLUTION
ln e
2x


1
2
This is the given expression.
 

d
ln e 2 x  1
dx
2
Differentiate.

 dxd ln e

e
1
d 2x

e 1
2x
 1 dx
Differentiate ln[g(x)].

e
1
 2e 2 x
2x
1
Finish.
2 ln e 2 x  1 
2 ln e 2 x  1 
2 ln e 2 x  1 
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2x

1

Use the power rule.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 43 of 55
Differentiating Logarithmic Expressions
EXAMPLE
The function f x  ln x  1 / x has a relative extreme point for x > 0. Find the
coordinates of the point. Is it a relative maximum point?
SOLUTION
f x  ln x  1 / x
This is the given function.
1
x   ln x  1 1
f x   x
x2
Use the quotient rule to
differentiate.
f  x   
ln x
x2
Simplify.
0
ln x
x2
Set the derivative equal to 0.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 44 of 55
Differentiating Logarithmic Expressions
CONTINUED
The derivative will equal 0 when the numerator equals 0 and the denominator
does not equal 0.
0  ln x
Set the numerator equal to 0.
x  e0  1
Write in exponential form.
To determine whether the function has a relative maximum at x = 1, let’s use the
second derivative.
f  x   
ln x
x2
1
 x 2   ln x   2 x 
x
f x  
 x 2 2
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This is the first derivative.
Differentiate.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 45 of 55
Differentiating Logarithmic Expressions
CONTINUED
f x  
 x  2 x ln x
x4
Simplify.
f  x  
 1  2 ln x
x3
Factor and cancel.
f 1 
 1  2 ln 1  1  2  0

 1
13
1
Evaluate the second derivative
at x = 1.
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
f 1  ln 1  1 / 1  0  1 / 1  1.
So, the relative maximum occurs at (1, 1).
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 46 of 55
§ 4.6
Properties of the Natural Logarithm Function
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 47 of 55
Section Outline

Properties of the Natural Logarithm Function

Simplifying Logarithmic Expressions

Differentiating Logarithmic Expressions

Logarithmic Differentiation
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 48 of 55
Properties of the Natural Logarithm Function
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 49 of 55
Simplifying Logarithmic Expressions
EXAMPLE
1
2
Write 5 ln x  ln y  3 ln z as a single logarithm.
SOLUTION
1
5 ln x  ln y  3 ln z
2
ln x 5  ln y1 2  ln z 3
x5
ln 1 2  ln z 3
y
This is the given expression.
Use LIV (this must be done
first).
Use LIII.
 x5 3 
ln  1 2 z 
y

Use LI.
 x5 z 3 
ln  1 2 
 y 
Simplify.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 50 of 55
Differentiating Logarithmic Expressions
EXAMPLE
Differentiate.
 x x  12 x  23 
ln 

4
x

1


SOLUTION
 x  x  12  x  23 
ln 

4
x

1


ln

This is the given expression.

x x  1 x  2  ln 4 x  1
2
3
Rewrite using LIII.
ln x  ln x  1  ln x  2  ln 4 x  1
Rewrite using LI.
1
ln x  2 ln  x  1  3 ln x  2   ln 4 x  1
2
Rewrite using LIV.
d 1







ln
x

2
ln
x

1

3
ln
x

2

ln
4
x

1

dx  2
Differentiate.
2
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3
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 51 of 55
Differentiating Logarithmic Expressions
CONTINUED
d 1
d
d
 d








ln 4 x  1
ln
x

2
ln
x

1

3
ln
x

2



dx  2
dx
dx
 dx
Distribute.
1 1
1
1
1
 2
3

4
2 x
x 1
x  2 4x 1
Finish differentiating.
1
2
3
4



2x x 1 x  2 4x 1
Simplify.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 52 of 55
Logarithmic Differentiation
Definition
Example
Logarithmic Differentiation:
Given a function y = f (x),
take the natural logarithm of
both sides of the equation,
Example will follow.
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.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 53 of 55
Logarithmic Differentiation
EXAMPLE
Use logarithmic differentiation to differentiate the function.
3
4

x  2 x  3
f x  
x  45
SOLUTION
3
4

x  2 x  3
f x  
x  45
This is the given function.
 x  2 3 x  34 
ln  f  x   ln 

5


x

4



3

3
Take the natural logarithm of
both sides of the equation.
 
ln  f x   ln x  2 x  3  ln x  4
4
 
 
5

Use LIII.
ln  f x   ln x  2  ln x  3  ln x  4
© 2010 Pearson Education Inc.
4
5

Use LI.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 54 of 55
Logarithmic Differentiation
CONTINUED
ln  f x  3 ln x  2  4 ln x  3  5 ln x  4

d
ln  f x   f x   3  4  5
dx
f x  x  2 x  3 x  4
4
5 
 3
f x   f x 



 x 2 x 3 x  4
3
4

x  2  x  3  3
4
5 
f  x  




x  45  x  2 x  3 x  4 
© 2010 Pearson Education Inc.
Use LIV.
Differentiate.
Solve for f ΄(x).
Substitute for f (x).
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e– Slide 55 of 55