LESSON X



LESSON 3 DERIVATIVES AND INTEGRALS INVOLVING THE

NATURAL AND GENERAL LOGARITHMIC FUNCTIONS

Definition The logarithmic function with base b is the function defined by [pic], where [pic] and [pic].

Recall that [pic] if and only if [pic]

Recall the following information about logarithmic functions:

1. The domain of [pic] is the set of positive real numbers. That is, the domain of [pic] is [pic].

2. The range of [pic] is the set of real numbers. That is, the range of [pic] is [pic].

3. The logarithmic function [pic] and the exponential function [pic] are inverses of one another. We studied the calculus of exponential functions in Lesson 2.

Definition The natural logarithmic function is the logarithmic function whose base is the irrational number e. Thus, the natural logarithmic function is the function defined by [pic], where [pic]. Recall that [pic].

Definition The common logarithmic function is the logarithmic function whose base is the number 10. Thus, the common logarithmic function is the function defined by [pic]. Recall that [pic].

Recall the following properties of logarithms:

1. [pic] = [pic]

2. [pic] = [pic] + [pic]

3. [pic] = [pic]

4. [pic]

5. [pic]

6. [pic]

7. [pic]

8. Change of Bases Formula: [pic]

Theorem If [pic], then [pic].

Proof Note that the domain of the function f is the set of real numbers x such that [pic]. Let [pic]. Then [pic]. If [pic], then [pic]. Thus,

[pic]. If [pic], then by the definition of logarithm, we have that [pic]. Using implicit differentiation, we obtain that [pic]. Solving for [pic], we have that [pic]. Since [pic], then [pic]. We have proven that if [pic] and [pic], then [pic]. Now, we want to prove that if [pic] and [pic], then [pic]. If [pic], then [pic]. Thus, [pic] and [pic]. Now, using the Chain Rule and the first result proved, we obtain that [pic]. Thus, [pic].

Theorem If [pic], where [pic], then [pic] or [pic].

Proof Use the Chain Rule and the fact that [pic].

Theorem If [pic], then [pic].

Proof Using the change of bases formula for logarithms, we have that [pic] = [pic]. Thus, [pic] =[pic]. Since [pic] is a constant and [pic], then [pic] = [pic].

Theorem If [pic], where [pic], then [pic] or [pic].

Proof Use the Chain Rule and the fact that [pic].

Examples Find the domain of the following function. Then differentiate them.

1. [pic]

We are taking the natural logarithm of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. Since the expression [pic] is positive when [pic], then the domain of the function f is the set of real numbers x such that [pic].

Since [pic] = [pic], then [pic].

Answer: Domain: [pic] or [pic]

[pic]

2. [pic]

We are taking the logarithm base 2 of the expression x. In order for the logarithm to be defined, we need the expression x to be positive. Thus, the domain of the function g is the set of positive real numbers.

[pic] [pic] =

[pic] = [pic]

Answer: Domain: [pic] or [pic]

[pic]

NOTE: The set of positive real numbers is also the domain of the function [pic].

3. [pic]

We are taking the natural logarithm of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. Since the expression [pic] is positive when [pic], then we need to find the sign of the expression [pic] = [pic].

Sign of [pic]: + [pic] +

( (

[pic] [pic]

Thus, the domain of the function s is the set of real numbers given by [pic].

Since [pic] = [pic], then [pic] = [pic].

Answer: Domain: [pic]

[pic]

NOTE: The set of real numbers given by [pic] is also the domain of the function [pic].

NOTE: If you wanted to find the sign of the function [pic], then you would want to know when [pic] and when [pic] is undefined.

[pic] [pic]. However, this number is not in the domain of s nor [pic].

[pic] undefined [pic] [pic][pic]

[pic]. These numbers are also not in the domain of s nor [pic].

Sign of [pic]: [pic] +

( (

[pic] [pic]

Thus, the function s is increasing on the interval [pic] and is decreasing on the interval [pic].

4. [pic]

We are taking the natural logarithm of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. The expression [pic] is positive when [pic]. Since [pic] = [pic], then [pic] when [pic] and [pic]. Thus, the domain of the function s is the set of real numbers given by [pic] or [pic].

Since [pic], then [pic] = [pic].

Answer: Domain: [pic] or

[pic]

[pic]

NOTE: If you wanted to find the sign of the function [pic], then you would want to know when [pic] and when [pic] is undefined.

[pic] [pic]. This number is now in the domain of s and [pic].

[pic] undefined [pic] [pic][pic]

[pic]. These numbers are still not in the domain of s nor [pic].

Sign of [pic]: [pic] + [pic] +

( ( (

[pic] [pic] [pic]

Thus, the function s is increasing on the set of real numbers given by [pic] and is decreasing on the set of real numbers given by [pic].

Since the numbers [pic] and [pic] are not in the domain of the function s, then the only critical number for the function s is [pic]. By the sign graph for [pic] above, there is a local maximum occurring when [pic] and the local maximum is [pic] = [pic] =

[pic] = [pic] = [pic] =

[pic].

5. [pic]

We are taking the logarithm base 3 of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. The expression [pic] is positive when [pic]. Then [pic] when [pic]. Thus, [pic] is a defined as a real number whenever [pic]. The square root of [pic] will be defined as a real number (not a complex number) if and only if [pic]. Since the base of the logarithm is three, which is greater than one, then [pic] if and only if [pic]. If the base of the logarithm would have been between 0 and 1, then [pic] if and only if [pic] and [pic]. Thus, the domain of the function y is the set of all real numbers x such that [pic]. In order to determine this set of real numbers, we need to solve the inequality [pic]. Since [pic], then we can solve the inequality [pic] . Thus, [pic] [pic] or [pic] [pic] or [pic]. Thus, the domain of the function y is the set of real numbers given by [pic].

Since [pic] = [pic], then

[pic] =

[pic] =

[pic] =

[pic] = [pic]

Answer: Domain: [pic]

[pic]

NOTE: The domain of the function y is the set of real numbers given by [pic] and the domain of the function [pic] is the set of real numbers given by [pic].

6. [pic]

Note: [pic]

We are taking the common (base 10) logarithm of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. Since the expression [pic] is nonnegative for all values of w, then we just need to see if the expression [pic] can equal zero. This expression is quadratic in [pic]. Since [pic] does not factor, we can use the quadratic formula to obtain the real solution(s) if they exist. Calculating the discriminant of this quadratic, we have that [pic] = [pic] . Thus, the four solutions to the quadratic equation [pic] are complex numbers. Thus, the expression [pic] can not equal zero for any real number w. Thus, the expression [pic] will be positive if the expression [pic] is positive. The expression [pic] is positive when [pic]. Solving this linear inequality, we have that [pic] when [pic]. Thus, the domain of the function y is the interval of real numbers given by [pic].

Since [pic]= [pic] +

[pic] = [pic] + [pic],

then [pic] =

[pic] + [pic]. Thus,

[pic] =

[pic] =

[pic]

Answer: Domain: [pic]

[pic]

7. [pic]

We are taking the logarithm base [pic] of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. Thus, we need to find the sign of the expression [pic].

Sign of [pic]: [pic] + [pic]

( (

[pic] [pic]

Thus, the domain of the function f is the set of real numbers given by [pic].

Since [pic] = [pic],

then [pic] =

[pic]

Answer: Domain: [pic]

[pic]

8. [pic]

We are taking the natural logarithm of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. Thus, we need to find the sign of the expression [pic].

Sign of [pic]:

[pic] + + [pic] +

( ( ( (

[pic] [pic] 0 [pic]

Thus, the expression [pic] is positive for the set of real numbers given by [pic]. Thus, the domain of the function h is the set of real numbers given by

[pic].

Since [pic] = [pic] ,

then [pic] = [pic].

Answer: Domain: [pic]

[pic]

9. [pic]

We are taking the natural logarithm of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. First of all, the expression [pic] is defined as long as [pic]. Thus, the expression [pic] is defined as long as [pic]. The defined expression [pic] is positive when [pic] [pic], [pic], and [pic]. Thus, the domain of the function h is the set of real numbers given by [pic] or

[pic].

Since [pic] = [pic] ,

then [pic] = [pic].

Answer: Domain: [pic]

or [pic]

[pic]

NOTE: We would do the following calculation in order to find [pic]:

[pic] = [pic] = [pic] =

[pic].

NOTE: This calculation could not have been done for the function h in Example 7 above because [pic] was not in the domain of that function.

10. [pic]

We are taking the logarithm base [pic] of the expression [pic]. In order for the logarithm to be defined, we need the expression [pic] to be positive. The expression [pic] will be positive when the expression [pic] is positive. Thus, we will need to find the sign of the expression [pic]. In order to do this, we need to find the solutions to the equations [pic] and [pic]. The four solutions to the equation [pic] are complex numbers. Since [pic] =

[pic] = [pic], then the real solutions to the equation [pic] are [pic] and [pic].

Sign of [pic]: + [pic] +

( (

[pic] 3

Thus, the domain of the function g is the set of real numbers given by [pic].

Since [pic] = [pic] =

[pic] = [pic] , then

[pic] =

[pic] =

[pic] =

[pic] = [pic]

Answer: Domain: [pic]

[pic]

NOTE: [pic]

11. [pic]

Domain of f = [pic]. The three solutions to the equation

[pic] can be visualized since they are the x-coordinate of the three points of intersection of the graphs of [pic] and [pic]. Of course, one of the solutions is [pic]. The other two solutions could be approximated using the Intermediate Value Theorem or Newton’s Method. The approximate value of these two solutions are [pic] and

[pic].

[pic] = [pic]

Answer: Domain: [pic]

[pic]

12. [pic]

Domain of y = [pic].

Since [pic] = [pic] = [pic], then

[pic] = [pic] = [pic] =

[pic]

Answer: Domain: [pic]

[pic]

13. [pic]

Since secant is the reciprocal of cosine, then the domain of y = [pic]. Since the solution to the equation [pic] is given by [pic], where n is an integer, then the solutions to the equation [pic] are given by [pic], where n is a nonnegative integer. Note that since the expression [pic] is negative when n is a negative integer. Since [pic] is positive, then n must be a

nonnegative integer. Since [pic] [pic], then [pic] = [pic] where n is a nonnegative integer[pic].

Since [pic] = [pic] , then

[pic] = [pic] =

[pic] = [pic]

Answer: Domain: [pic] where n is a nonnegative

integer[pic]

[pic]

14. [pic]

The domain of g = [pic]. Thus, we need to solve the nonlinear inequality [pic] . Since the expression [pic] is quadratic in [pic] and factors. We have that [pic] = [pic]. Since the expression [pic] is positive for all real numbers x, then we need that the factor [pic] to be positive in order for [pic]. Thus, we just need to solve the nonlinear inequality [pic]. We have that [pic] [pic] . Since the function [pic] is an increasing function for all positive real numbers, then [pic] [pic] [pic] [pic] [pic].

Thus, the domain of g = [pic] = [pic] = [pic].

Since [pic], then [pic] =

[pic]

Answer: Domain: [pic] or [pic]

[pic]

15. [pic]

[pic] [pic]

[pic]

[pic]

[pic]

[pic]

Answer: [pic]

16. [pic]

[pic] [pic]

[pic] [pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Answer: [pic]

17. [pic]

[pic] [pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Examples Use logarithmic differentiation in order to differentiate the following functions.

1. [pic]

[pic] [pic]

[pic]

[pic]

[pic]

[pic] =

[pic]

Answer: [pic]

NOTE: This answer would not allow us to find the sign of [pic], but it would allow us to find the value of [pic] at any number in the domain of [pic]. For example,

[pic] = [pic] =

[pic] = [pic] = [pic].

2. [pic]

[pic] [pic] [pic]

[pic] [pic]

[pic] =

[pic]

Answer: [pic] OR

[pic]

3. [pic]

[pic] [pic]

[pic]

[pic]

[pic]

[pic] =

[pic] =

[pic]

Answer: [pic] OR

[pic]

4. [pic]

[pic] [pic] [pic]

[pic]

[pic]

[pic]

[pic] =

[pic]

Answer: [pic]

5. [pic]

[pic] [pic] [pic]

[pic]

[pic]

[pic] =

[pic] =

[pic] =

[pic] =

[pic]

Answer: [pic]

Examples Determine the interval(s) on which the following functions are increasing and decreasing, the interval(s) on which they are concave upward and concave downward. Also, find their local maximum(s), local minimum(s), and inflection point(s).

1. [pic]

The domain of the function f is the interval [pic].

[pic] = [pic] =

[pic] = [pic] =

[pic]

[pic] [pic]

[pic] [pic] [pic] [pic]

[pic] [pic]

Sign of [pic]: [pic] +

( (

0 [pic]

The function f is increasing on the interval [pic] and is decreasing on the interval [pic].

The function f has a local minimum occurring when [pic] and the local minimum is [pic]. Since [pic] = [pic], then [pic] = [pic] = [pic] =

[pic] = [pic] =

[pic] = [pic] = [pic]

Sign of [pic]: +

(

0

The function f is concave upward on the interval [pic] and is concave downward nowhere.

The function f has no inflection point.

Answer: Increasing: [pic] Decreasing: [pic]

Local Maximum: None Local Minimum: [pic]

CU: [pic] CD: Nowhere

Inflection Point: None

Since [pic], then [pic].

Examples Evaluate the following integrals.

1. [pic]

The integrand [pic] is continuous on the closed interval [pic]. It only has a discontinuity at [pic]. Thus, we can apply the Fundamental Theorem of Calculus.

[pic]

Answer: [pic]

2. [pic]

Let [pic]

Then [pic]

[pic] = [pic] = [pic] = [pic] =

[pic] = [pic]

Answer: [pic]

3. [pic]

The integrand [pic] is continuous on the closed interval [pic]. It only has discontinuities at [pic] and [pic]. Thus, we can apply the Fundamental Theorem of Calculus.

Let [pic]

Then [pic]

[pic] = [pic] = [pic] = [pic] =

[pic] = [pic] = [pic] = [pic]

NOTE: Since [pic], then when [pic], we have that [pic]. Thus, the new lower limit of integration is [pic]. Similarly, when [pic], we have that [pic]. Thus, the new upper limit of integration is [pic].

Answer: [pic]

4. [pic]

[pic] = [pic] =

[pic]

Answer: [pic]

5. [pic]

The integrand [pic] is continuous on the closed interval [pic]. It only has a discontinuity at [pic]. Thus, we can apply the Fundamental Theorem of Calculus.

[pic] = [pic] = [pic] =

[pic] = [pic] =

[pic] = [pic] =

[pic] = [pic] = [pic] =

[pic] = [pic] = [pic]

Answer: [pic]

6. [pic]

Let [pic]

Then [pic]

[pic] = [pic] = [pic] = [pic] = [pic] =

[pic]

Answer: [pic]

7. [pic]

The integrand [pic] is continuous on the closed interval [pic]. It only has a discontinuity at [pic]. Thus, we can apply the Fundamental Theorem of Calculus.

Let [pic]

Then [pic]

[pic] = [pic] = [pic] =

[pic] = [pic] = [pic] = [pic] =

[pic] = [pic] = [pic]

NOTE: Since [pic], then when [pic], we have that [pic]. Thus, the new lower limit of integration is 1. Similarly, when [pic], we have that [pic]. Thus, the new upper limit of integration is 3.

Answer: [pic]

8. [pic]

Let [pic]

Then [pic]

[pic] = [pic] = [pic] = [pic] =

[pic]

Answer: [pic]

9. [pic]

Let [pic]

Then [pic]

[pic] = [pic] = [pic] = [pic] =

[pic] = [pic] = [pic]

Answer: [pic]

10. [pic]

The integrand [pic] is continuous on the closed interval [pic]. It does not have any discontinuities. Thus, we can apply the Fundamental Theorem of Calculus.

Let [pic]

Then [pic] = [pic]

[pic] = [pic] = [pic]

= [pic] = [pic] = [pic] =

[pic]

NOTE: Since [pic], then when [pic], we have that [pic] = [pic]. Thus, the new lower limit of integration is [pic]. Similarly, when [pic], we have that [pic] = [pic] = [pic] [pic]. Thus, the new upper limit of integration is [pic].

Answer: [pic]

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