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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 55
Chapter 4
The Exponential and Natural Logarithm
Functions
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 55
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 55
§ 4.1
Exponential Functions
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 55
Exponential Functions
Properties of Exponential Functions
Simplifying Exponential Expressions
Graphs of Exponential Functions
Solving Exponential Equations
Section Outline
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 55
Exponential Function
Definition Example
Exponential Function: A function whose exponent is the independent variable
xy 3
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 55
Properties of Exponential Functions
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 55
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 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.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 55
Solving Exponential Equations
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.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 55
§ 4.2
The Exponential Function e x
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 55
e
The Derivatives of 2x, bx, and ex
Section Outline
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 55
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 55
The Derivative of 2x
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 55
Solving Exponential Equations
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Calculate.
2
2x
x
dx
d
173.025.0693.02 2 2
2
mdx
d
x
x
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 15 of 55
The Derivatives of bx and ex
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 16 of 55
Solving Exponential Equations
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.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 17 of 55
Solving Exponential Equations
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 18 of 55
Solving Exponential Equations
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.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 19 of 55
§ 4.3
Differentiation of Exponential Functions
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 20 of 55
Chain Rule for eg(x)
Working With Differential Equations
Solving Differential Equations at Initial Values
Functions of the form ekx
Section Outline
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 21 of 55
Chain Rule for eg(x)
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 22 of 55
Chain Rule for eg(x)
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 23 of 55
Working With Differential Equations
Generally speaking, a differential equation is an equation that contains a derivative.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 24 of 55
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.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 25 of 55
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 26 of 55
Functions of the form ekx
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 27 of 55
§ 4.4
The Natural Logarithm Function
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 28 of 55
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
Section Outline
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 29 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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 30 of 55
Properties of the Natural Logarithm
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 31 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.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 32 of 55
Exponential Expressions
EXAMPLEEXAMPLE
SOLUTIONSOLUTION
Simplify.xe ln23ln
Using properties of the exponential function, we have
.3332ln2ln2ln2
3lnln23ln
xeeeee
ee
xxxx
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 33 of 55
Solving Exponential Equations
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 34 of 55
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 35 of 55
Other Exponential and Logarithmic Functions
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 36 of 55
Common Logarithms
Definition Example
Common logarithm: Logarithms to the base 10
2100log10
31000log10
4000,10log10
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 37 of 55
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 38 of 55
Max’s & Min’s of Exponential Equations
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.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 39 of 55
Max’s & Min’s of Exponential Equations
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).
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 40 of 55
§ 4.5
The Derivative of ln x
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 41 of 55
Derivatives for Natural Logarithms
Differentiating Logarithmic Expressions
Section Outline
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 42 of 55
Derivative Rules for Natural Logarithms
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 43 of 55
Differentiating Logarithmic Expressions
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 44 of 55
Differentiating Logarithmic Expressions
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 45 of 55
Differentiating Logarithmic Expressions
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 46 of 55
Differentiating Logarithmic Expressions
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).
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 47 of 55
§ 4.6
Properties of the Natural Logarithm Function
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 48 of 55
Properties of the Natural Logarithm Function
Simplifying Logarithmic Expressions
Differentiating Logarithmic Expressions
Logarithmic Differentiation
Section Outline
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 49 of 55
Properties of the Natural Logarithm Function
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 50 of 55
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 51 of 55
Differentiating Logarithmic Expressions
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 52 of 55
Differentiating Logarithmic Expressions
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 53 of 55
Logarithmic Differentiation
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.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 54 of 55
Logarithmic Differentiation
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
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 55 of 55
Logarithmic Differentiation
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