Big Idea 2: Derivative. Differentiation - Review the Knowledge You Need to Score High - 5 Steps to a 5: AP Calculus AB 2017 (2016)

5 Steps to a 5: AP Calculus AB 2017 (2016)

STEP 4

Review the Knowledge You Need to Score High

CHAPTER 7

Big Idea 2: Derivatives

Differentiation

IN THIS CHAPTER

Summary: The derivative of a function is often used to find rates of change. It is also related to the slope of a tangent line. On the AP Calculus AB exam, many questions involve finding the derivative of a function. In this chapter, you will learn different techniques for finding a derivative which include using the Power Rule, Product and Quotient Rules, Chain Rule, and Implicit Differentiation. You will also learn to find the derivatives of trigonometric, exponential, logarithmic, and inverse functions.

Images

Key Ideas

Images Definition of the derivative of a function

Images Power Rule, Product and Quotient Rules, and Chain Rule

Images Derivatives of trigonometric, exponential, and logarithmic functions

Images Derivatives of inverse functions

Images Implicit Differentiation

Images Higher order derivatives

Images L’Hôpital’s Rule for Indeterminate Forms


7.1 Derivatives of Algebraic Functions

Main Concepts: Definition of the Derivative of a Function; Power Rule; The Sum, Difference, Product, and Quotient Rules; The Chain Rule

Definition of the Derivative of a Function

The derivative of a function f , written as f ′, is defined as

Images

if this limit exists. (Note that f ′(x ) is read as f prime of x .)

Other symbols of the derivative of a function are:

Images .

Let m tangent be the slope of the tangent to a curve y = f (x ) at a point on the curve. Then,

Images .

(See Figure 7.1-1 .)

Images

Figure 7.1-1

Given a function f , if f ′(x ) exists at x = a , then the function f is said to be differentiable at x = a . If a function f is differentiable at x = a , then f is continuous at x = a . (Note that the converse of the statement is not necessarily true, i.e., if a function f is continuous at x = a , then f may or may not be differentiable at x = a .) Here are several examples of functions that are not differentiable at a given number x = a . (See Figures 7.1-27.1-5 .)

Images

Figure 7.1-2

Images

Figure 7.1-3

Images

Figure 7.1-4

Images

Figure 7.1-5

Example 1

If f (x ) = x 2 – 2x – 3, find (a) f ′(x ) using the definition of derivative, (b) f ′(0), (c) f ′(1), and (d) f ′(3).

(a) Using the definition of derivative, Images

Images

(b) f ′(0) = 2(0) – 2 = –2, (c) f ′(1) = 2(1) – 2 = 0, and (d) f ′(3) = 2(3) – 2 = 4.

Example 2

Evaluate Images .

The expression Images is equivalent to the derivative of the function f (x ) = cos x at x = π , i.e., f ′(π ). The derivative of f (x ) = cos x at x = π is equivalent to the slope of the tangent to the curve of cos x at x= π . The tangent is parallel to the x -axis. Thus, the slope is 0 or Images = 0.

Or, using an algebraic method, note that cos(a + b ) = cos(a ) cos(b ) – sin(a ) sin(b ). Then rewrite ImagesImages .

(See Figure 7.1-6 .)

Images

Figure 7.1-6

Example 3

If the function f (x ) = x 2/3 + 1, find all points where f is not differentiable.

The function f (x ) is continuous for all real numbers and the graph of f (x ) forms a “cusp” at the point (0, 1). Thus, f (x ) is not differentiable at x = 0. (See Figure 7.1-7 .)

Images

Figure 7.1-7

Example 4

Using a calculator, find the derivative of f (x ) = x 2 + 4x at x = 3.

There are several ways to find f ′(3), using a calculator. One way is to use the [nDeriv ] function of the calculator. From the main Home screen, select F3-Calc and then select [nDeriv ]. Enter [nDeriv ] (x 2 + 4x , x )|x = 3. The result is 10.

Images

• Always write out all formulas in your solutions.

Power Rule

If f (x ) = c where c is a constant, then f ′(x ) = 0.

If f (x ) = xn where n is a real number, then f ′(x ) = nx n –1 .

If f (x ) = c x n where c is a constant and n is a real number, then f ′(x ) = cnx n –1 .

Summary of Derivatives of Algebraic Functions

Images

Example 1

If f (x ) = 2x 3 , find (a) f ′(x ), (b) f ′(1), and (c) f ′(0).

Note that (a) f ′(x ) = 6x 2 , (b) f ′(1) = 6(1)2 = 6, and (c) f ′(0) = 0.

Example 2

If Images , find (a) Images and (b) Images (which represents Images at x = 0).

Note that (a) Images and thus, Images and (b) Images does not exist because the expression Images is undefined.

Example 3

Here are several examples of algebraic functions and their derivatives:

Images

Example 4

Using a calculator, find f ′(x ) and f ′(3) if Images .

There are several ways of finding f ′(x ) and f ′(9) using a calculator. One way is to use the d [Differentiate ] function. Go to the Home screen. Select F3-Calc and then select d [Differentiate ]. Enter Images . The result is Images . To find f ′(3), enter Images x )|x = 3. The result is Images .

Images

The Sum, Difference, Product, and Quotient Rules

If u and v are two differentiable functions, then

Images

Example 1

Find f ′(x ) if f (x ) = x 3 – 10x + 5.

Using the sum and difference rules, you can differentiate each term and obtain f ′(x ) = 3x 2 – 10. Or using your calculator, select the d [Differentiate ] function and enter d (x 3 – 10x + 5, x ) and obtain 3x 2 – 10.

Example 2

If y = (3x – 5)(x 4 + 8x – 1), find Images .

Using the product rule Images , let u = (3x – 5) and v = (x 4 + 8x – 1).

Then Images = (3)(x 4 + 8x – 1) + (4x 3 + 8)(3x – 5) = (3x 4 + 24x – 3) + (12x 4 – 20x 3 + 24x – 40) = 15x 4 – 20x 3 + 48x – 43. Or you can use your calculator and enter d ((3x – 5)(x 4 + 8x – 1), x ) and obtain the same result.

Example 3

If Images , find f ′(x ).

Using the quotient rule Images , let u = 2x – 1 and v = x + 5. Then Images . Or you can use your calculator and enter d ((2x – 1)/(x + 5), x ) and obtain the same result.

Example 4

Using your calculator, find an equation of the tangent to the curve f (x ) = x 2 – 3x + 2 at x = 5.

Find the slope of the tangent to the curve at x = 5 by entering d (x 2 – 3x + 2, x )|x = 5. The result is 7. Compute f (5) = 12. Thus, the point (5, 12) is on the curve of f (x ). An equation of the line whose slope m = 7 and passing through the point (5, 12) is y – 12 = 7(x – 5).

Images

• Remember that Images ln Images and Images . The integral formula is not usually tested in the AB exam.

The Chain Rule

If y = f (u ) and u = g (x ) are differentiable functions of u and x respectively, then Images or Images .

Example 1

If y = (3x – 5)10 , find Images .

Using the chain rule, let u = 3x – 5 and thus, y = u 10 . Then, Images and Images .

Since Images (3) = 10(3x – 5)9 (3) = 30(3x – 5)9 . Or you can use your calculator and enter d ((3x – 5)10 , x ) and obtain the same result.

Example 2

If Images , find f ′(x ).

Rewrite Images as f (x ) = 5x (25 – x 2 )1/2. Using the product rule, f ′(x ) = Images .

To find Images , use the chain rule and let u = 25 – x 2 .

Thus, Images . Substituting this quantity back into f ′(x ), you have Images Images . Or you can use your calculator and enter Images and obtain the same result.

Example 3

If Images , find Images .

Using the chain rule, let Images . Then Images .

To find Images , use the quotient rule.

Thus, Images . Substituting this quantity back into Images .

An alternate solution is to use the product rule and rewrite Images as y = Images and use the quotient rule. Another approach is to express y = (2x – 1)3 (x –6 ) and use the product rule. Of course, you can always use your calculator if you are permitted to do so.

7.2 Derivatives of Trigonometric, Inverse Trigonometric, Exponential, and Logarithmic Functions

Main Concepts: Derivatives of Trigonometric Functions, Derivatives of Inverse Trigonometric Functions, Derivatives of Exponential and Logarithmic Functions

Derivatives of Trigonometric Functions
Summary of Derivatives of Trigonometric Functions

Images

Note that the derivatives of cosine , cotangent , and cosecant all have a negative sign.

Example 1

If y = 6x 2 + 3 sec x , find Images .

Images sec x tan x .

Example 2

Find f ′(x ) if f (x ) = cot(4x – 6).

Using the chain rule, let u = 4x – 6. Then f ′(x ) = [– csc2 (4x – 6)][4] = –4 csc2 (4x – 6).

Or using your calculator, enter d (1/tan(4x – 6), x ) and obtain Images which is an equivalent form.

Example 3

Find f ′(x ) if f (x ) = 8 sin(x 2 ).

Using the chain rule, let u = x 2 . Then f ′(x ) = [8 cos(x 2 )][2x ] = 16x cos(x 2 ).

Example 4

If y = sin x cos(2x ), find Images .

Using the product rule, let u = sin x and v = cos(2x ).

Then Images .

Example 5

If y = sin[cos(2x )], find Images .

Using the chain rule, let u = cos(2x ). Then

Images .

To evaluate Images , use the chain rule again by making another u -substitution, this time for 2x . Thus, Images . Therefore, Images .

Example 6

Find f ′(x ) if f (x ) = 5x csc x .

Using the product rule, let u = 5x and v = csc x . Then f ′(x ) = 5 csc x + (– csc x cot x ) (5x ) = 5 csc x – 5x (csc x )(cot x ).

Example 7

If Images , find Images .

Rewrite Images as y = (sin x ) 1/2 . Using the chain rule, let u = sin x . Thus, Images Images .

Example 8

If Images , find Images .

Using the quotient rule, let u = tan x and v = (1 + tan x ). Then,

Images

Note: For all of the above exercises, you can find the derivatives by using a calculator, provided that you are permitted to do so.

Derivatives of Inverse Trigonometric Functions
Summary of Derivatives of Inverse Trigonometric Functions

Let u be a differentiable function of x , then

Images

Note that the derivatives of cos–1 x , cot–1 x , and csc–1 x all have a “–1” in their numerators.

Example 1

If y = 5 sin–1 (3x ), find Images .

Let u = 3x . Then Images .

Or using a calculator, enter d [5 sin–1 (3x ), x ] and obtain the same result.

Example 2

Find f ′(x ) Images .

Let Images

Example 3

If y = sec–1 (3x 2 ), find Images .

Let u = 3x 2 . Then Images .

Example 4

If Images .

Note: For all of the above exercises, you can find the derivatives by using a calculator, provided that you are permitted to do so.

Derivatives of Exponential and Logarithmic Functions
Summary of Derivatives of Exponential and Logarithmic Functions

Let u be a differentiable function of x , then

Images .

For the following examples, find Images and verify your result with a calculator.

Example 1

y = e 3x + 5xe 3 + e 3

Images (Note that e 3 is a constant.)

Example 2

y = xex x 2 ex

Using the product rule for both terms, you have

Images

Example 3

y = 3sin x

Let u = sin x . Then, Images .

Example 4

y = e (x 3 )

Let u = x 3 . Then, Images .

Example 5

y = (ln x )5

Let u = ln x . Then, Images .

Example 6

y = ln(x 2 + 2x – 3) + ln 5

Let u = x 2 + 2x – 3. Then, Images .

(Note that ln 5 is a constant. Thus the derivative of ln 5 is 0.)

Example 7

y = 2x ln x + x

Using the product rule for the first term,

you have Images (2x ) + 1 = 2 ln x + 2 + 1 = 2 ln x + 3.

Example 8

y = ln(ln x )

Let u = ln x . Then Images .

Example 9

y = log5 (2x + 1)

Let u = 2x + 1. Then Images .

Example 10

Write an equation of the line tangent to the curve of y = ex at x = 1.

The slope of the tangent to the curve y = ex at x = 1 is equivalent to the value of the derivative of y = e x evaluated at x = 1. Using your calculator, enter d (e (x ), x )|x = 1 and obtain e . Thus, m = e , the slope of the tangent to the curve at x = 1. At x = 1, y = e 1 = e , and thus the point on the curve is (1, e ). Therefore, the equation of the tangent is ye = e (x – 1) or y = ex . (See Figure 7.2-1 .)

Images

Figure 7.2-1

Images

• Never leave a multiple-choice question blank. There is no penalty for incorrect answers.

7.3 Implicit Differentiation

Main Concept: Procedure for Implicit Differentiation

Images

Procedure for Implicit Differentiation

Given an equation containing the variables x and y for which you cannot easily solve for y in terms of x , you can find Images by doing the following:

Steps

1: Differentiate each term of the equation with respect to x .

2: Move all terms containing Images to the left side of the equation and all other terms to the right side.

3: Factor out Images on the left side of the equation.

4: Solve for Images .

Example 1

Find Images if y 2 – 7y + x 2 – 4x = 10.

Step 1: Differentiate each term of the equation with respect to x . (Note that y is treated as a function of x .) Images

Step 2: Move all terms containing Images to the left side of the equation and all other terms to the right: Images .

Step 3: Factor out Images .

Step 4: Solve for Images .

Example 2

Given x 3 + y 3 = 6xy , find Images .

Step 1: Differentiate each term with respect to Images .

Step 2: Move all Images terms to the left side: Images .

Step 3: Factor out Images .

Step 4: Solve for Images .

Example 3

Find Images if (x + y ) 2 – (xy ) 2 = x 5 + y 5 .

Step 1: Differentiate each term with respect to x :

Images

Distributing 2(x + y ) and –2(xy ), you have.

Images

Step 2: Move all Images terms to the left side:

Images

Step 3: Factor out Images :

Images

Step 4: Solve for Images .

Example 4

Write an equation of the tangent to the curve x 2 + y 2 + 19 = 2x + 12y at (4, 3). The slope of the tangent to the curve at (4, 3) is equivalent to the derivative Images at (4, 3).

Using implicit differentiation, you have:

Images

Thus, the equation of the tangent is y – 3 = (1)(x – 4) or y – 3 = x – 4.

Example 5

Find Images , if sin(x + y ) = 2x .

Images .

7.4 Approximating a Derivative

Given a continuous and differentiable function, you can find the approximate value of a derivative at a given point numerically. Here are two examples.

Example 1

The graph of a function f on [0, 5] is shown in Figure 7.4-1 . Find the approximate value of f ′(3).

Images

Figure 7.4-1

Since f ′(3) is equivalent to the slope of the tangent to f (x ) at x = 3, there are several ways you can find its approximate value.

Method 1: Use the slope of the line segment joining the points at x = 3 and x = 4.

Images

Method 2: Use the slope of the line segment joining the points at x = 2 and x = 3.

Images

Method 3: Use the slope of the line segment joining the points at x = 2 and x = 4.

Images

Note that Images is the average of the results from methods 1 and 2.

Thus, f ′(3) ≈ 1, 2, or Images depending on which line segment you use.

Example 2

Let f be a continuous and differentiable function. Selected values of f are shown below. Find the approximate value of f ′ at x = 1.

Images

You can use the difference quotient Images to approximate f ′(a ).

Images

Or, you can use the symmetric difference quotient Images to approximate f ′(a ).

Images

Thus, f ′(3) ≈ 0.49, 0.465, 0.54, or 0.63 depending on your method.

Note that f is decreasing on (–2, –1) and increasing on (–1, 3). Using the symmetric difference quotient with h = 3 would not be accurate. (See Figure 7.4-2 .)

Images

Figure 7.4-2

Images

• Remember that the Images because the Images .

7.5 Derivatives of Inverse Functions

Let f be a one-to-one differentiable function with inverse function f –1 . If f ′(f –1 (a )) ≠ 0, then the inverse function f –1 is differentiable at a and (f –1 )′(a ) = Images . (See Figure 7.5-1 .)

Images

Figure 7.5-1

If y = f –1 (x ) so that x = f (y ), then Images with Images .

Example 1

If f (x ) = x 3 + 2x – 10, find (f –1 )′(x ).

Step 1: Check if (f –1 )′(x ) exists. f ′(x ) = 3x 2 + 2 and f ′(x ) > 0 for all real values of x . Thus, f (x ) is strictly increasing which implies that f (x ) is 1 – 1. Therefore, (f –1 )′(x ) exists.

Step 2: Let y = f (x ) and thus y = x 3 + 2x – 10.

Step 3: Interchange x and y to obtain the inverse function x = y 3 + 2y – 10.

Step 4: Differentiate with respect to Images .

Step 5: Apply formula Images

Example 2

Example 1 could have been done by using implicit differentiation.

Step 1: Let y = f (x ), and thus y = x 3 + 2x – 10.

Step 2: Interchange x and y to obtain the inverse function x = y 3 + 2y – 10.

Step 3: Differentiate each term implicitly with respect to x .

Images

Step 4: Solve for Images .

Images

Example 3

If f (x ) = 2x 5 + x 3 + 1, find (a) f ′(1) and f ′(1) and (b) (f –1 )(4) and (f –1 )′(4).

Enter y 1 = 2x 5 + x 3 + 1. Since y 1 is strictly increasing, f (x ) has an inverse.

(a) f (1) = 2(1)5 + (1)3 + 1 = 4
f ′(x ) = 10x 4 + 3x 2
f ′(1) = 10(1)4 + 3(1)2 = 13

(b) Since f (1) = 4 implies the point (1, 4) is on the curve f (x ) = 2x 5 + x 3 + 1, therefore, the point (4, 1) (which is the reflection of (1, 4) on y = x ) is on the curve (f –1 )(x ). Thus, (f –1 )(4) = 1.

Images

Example 4

If f (x ) = 5x 3 + x + 8, find (f –1 )′(8).

Enter y 1 = 5x 3 + x + 8. Since y 1 is strictly increasing near x = 8, f (x ) has an inverse near x = 8.

Note that f (0) = 5(0)3 + 0 + 8 = 8 which implies the point (0, 8) is on the curve of f (x ).

Thus, the point (8, 0) is on the curve of (f –1 )(x ).

Images

Therefore, Images

Images

• You do not have to answer every question correctly to get a 5 on the AP Calculus AB exam. But always select an answer to a multiple-choice question. There is no penalty for incorrect answers.

7.6 Higher Order Derivatives

If the derivative f ′ of a function f is differentiable, then the derivative of f ′ is the second derivative of f represented by f ″ (reads as f double prime). You can continue to differentiate f as long as there is differentiability.

Some of the Symbols of Higher Order Derivatives

Images

Note that Images .

Example 1

If y = 5x 3 + 7x – 10, find the first four derivatives.

Images

Example 2

If Images , find f ″ (4).

Rewrite: Images and differentiate: Images .

Differentiate again:

Images

Example 3

If y = x cos x , find y ″.

Images

Or, you can use a calculator and enter d [x cos x , x , 2] and obtain the same result.

7.7 L’Hôpital’s Rule for Indeterminate Forms

Let lim represent one of the limits: Images or Images . Suppose f (x ) and g (x ) are differentiable and g ′(x ) = 0 near c , except possibly at c , and suppose lim f (x ) = 0 and lim g (x ) = 0. Then the Images is an indeterminate form of the type Images . Also, if lim f (x ) = ±∞ and lim g (x ) = ±∞, then the Images , is an indeterminate form of the type Images . In both cases, Images and Images , L’Hôpital’s Rule states that lim = lim Images .

Example 1

Find lim Images , if it exists.

Since Images Images , this limit is an indeterminate form. Taking the derivatives, Images and Images . By L’Hôpital’s Rule, Images Images .

Example 2

Find Images , if it exists.

Rewriting Images as Images shows that the limit is an indeterminate form, since Images and Images . Differentiating and applying L’Hôpital’s Rule means that Images . Unfortunately, this new limit is also indeterminate. However, it is possible to apply L’Hôpital’s Rule again, so Images equals to Images . This expression approaches zero as x becomes large, so Images .

7.8 Rapid Review

1. If y = ex 3 , find Images .

Answer: Using the chain rule, Images .

2. Evaluate Images .

Answer: The limit is equivalent to Images .

3. Find f ′(x ) if f (x ) = ln(3x ).

Answer: Images .

4. Find the approximate value of f ′(3). (See Figure 7.8-1 .)

Images

Figure 7.8-1

Answer: Using the slope of the line segment joining (2, 1) and (4, 3), Images .

5. Find Images if xy = 5x 2 .

Answer: Using implicit differentiation, Images . Thus, Images .

Or simply solve for y leading to y = 5x and thus, Images .

6. If Images , find Images .

Answer: Rewrite y = 5x - 2. Then, = Images and Images .

7. Using a calculator, write an equation of the line tangent to the graph f (x ) = –2x 4 at the point where f ′(x ) = –1.

Answer: f ′(x ) = –8x 3 . Using a calculator, enter [Solve ] [–8x 3 =-1, x ] and obtain Images . Using the calculator Images . Thus, tangent is Images .

8. Images

Answer : Since Images consider Images .

9. Images

Answer : Since Images consider Images .

7.9 Practice Problems

Part A The use of a calculator is not allowed.

Find the derivative of each of the following functions.

1 . y = 6x 5x + 10

2 . Images

3 . Images

4 . Images

5 . f (x ) = (3x – 2)5 (x 2 – 1)

6 . Images

7 . y = 10 cot(2x – 1)

8 . y = 3x sec(3x )

9 . y = 10 cos[sin(x 2 – 4)]

10 . y = 8 cos–1 (2x )

11 . y = 3e 5 + 4xe x

12 . y = ln(x 2 + 3)

Part B Calculators are allowed.

13 . Find Images , if x 2 + y 3 = 10 – 5xy .

14 . The graph of a function f on [1, 5] is shown in Figure 7.9-1 . Find the approximate value of f ′(4).

Images

Figure 7.9-1

15 . Let f be a continuous and differentiable function. Selected values of f are shown below. Find the approximate value of f ′ at x = 2.

Images

16 . If f (x ) = x 5 + 3x – 8, find (f –1 )′(–8).

17 . Write an equation of the tangent to the curve y = ln x at x = e .

18 . If y = 2x sin x , find Images at Images .

19 . If the function f (x ) = (x – 1)2 /3 + 2, find all points where f is not differentiable.

20 . Write an equation of the normal line to the curve x cos y = 1 at Images .

21 . Images

22 . Images

23 . Images

24 . Images

25 . Images

7.10 Cumulative Review Problems

(Calculator) indicates that calculators are permitted.

26 . Find Images

27 . If f (x ) = cos2 (πx ), find f ′(0).

28 . Find Images .

29 . (Calculator) Let f be a continuous and differentiable function. Selected values of f are shown below. Find the approximate value of f ′ at x = 2.

Images

30 . (Calculator) If Images determine if f (x ) is continuous at (x = 3). Explain why or why not.

7.11 Solutions to Practice Problems

Part A The use of a calculator is not allowed.

1 . Applying the power rule, Images .

2 . Rewrite Images as f (x ) = x –1 + x –2/3 . Differentiate: Images .

3 . Rewrite Images .
Differentiate: Images .
An alternate method is to differentiate Images directly, using the quotient rule.

4 . Applying the quotient rule,

Images

5 . Applying the product rule, u = (3x – 2)5 and v = (x 2 – 1), and then the chain rule,

Images

6 . Rewrite Images as.

Applying first the chain rule and then the quotient rule,

Images

Note: Images ,

if Images which implies Images or Images .

Another method is to write Images and use the product rule.

Another method is to write y = (2x + 1)1/2 (2x – 1)1/2 and use the product rule.

7 . Let u = 2x – 1,
Images

8 . Using the product rule,
Images

9 . Using the chain rule, let u = sin(x 2 – 4).

Images

10 . Using the chain rule, let u = 2x .

Images

11 . Since 3e 5 is a constant, its derivative is 0.

Images

12 . Let Images

Part B Calculators are allowed.

13 . Using implicit differentiation, differentiate each term with respect to x .

Images

14 . Since f ′(4) is equivalent to the slope of the tangent to f (x ) at x = 4, there are several ways you can find its approximate value.

Method 1: Use the slope of the line segment joining the points at x = 4 and x = 5.

Images

Method 2: Use the slope of the line segment joining the points at x = 3 and x = 4.

Images

Method 3: Use the slope of the line segment joining the points at x = 3 and x = 5.

Images

Note that –2 is the average of the results from methods 1 and 2. Thus f ′(4) ≈ –3, –1, or –2 depending on which line segment you use.

15 . You can use the difference quotient Images to approximate f ′(a ). Let h = 1; Images Images .

Or, you can use the symmetric difference quotient Images to approximate f ′(a ).

Let h = 1; Images Images .

Thus, f ′(2) ≈ 4 or 5 depending on your method.

16 . Enter y 1 = x 5 + 3x – 8. The graph of y 1 is strictly increasing. Thus f (x ) has an inverse. Note that f (0) = –8. Thus the point (0, –8) is on the graph of f (x ) which implies that the point (–8, 0) is on the graph of f 1(x ).
f ′(x ) = 5x 4 + 3 and f ′(0) = 3.

Since Images , thus Images .

17 . Images and Images Thus the slope of the tangent to y = ln x at x = e is Images . At x = e , y = ln x = ln e = 1, which means the point (e ,1) is on the curve of y = ln x . Therefore, an equation of the tangent is Imagesor Images See Figure 7.11-1 .

Images

Figure 7.11-1

18 . Images

Images

Or, using a calculator, enter d (2x – sin(x ), x , 2) Images and obtain –π .

19 . Enter y 1 = (x – 1)2 /3 + 2 in your calculator. The graph of y 1 forms a cusp at x = 1. Therefore, f is not differentiable at x = 1.

20 . Differentiate with respect to x :

Images

Thus, the slope of the tangent to the curve at (2, π /3) is Images The slope of the normal line to the curve at (2, π /3) is Images . Therefore an equation of the normal line is Images .

21 . Images

22 . Images

23 . Images

24 . Images

25 . Images

7.12 Solutions to Cumulative Review Problems

26 . The expression Images the derivative of sin x at x = π /2 which is the slope of the tangent to sin x at x = π /2. The tangent to sin x at x = π /2 is parallel to the x -axis.
Therefore the slope is 0, i.e., Images . An alternate method is to expand sin Images .

Images

27 . Using the chain rule, let u = (πx ).

Then, Images

28 . Since the degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator, the limit is 0.

29 . You can use the differencequotient Images to approximate f ′(a ).

Images

Or, you can use the symmetric difference quotient Images to approximate f ′(a ).

Images

Thus, f ′(2) = 1.7, 2.05, or 1.25 depending on your method.

30 . (See Figure 7.12–1 .) Checking the three conditions of continuity:

Images

Figure 7.12-1

(1) f (3) = 3

(2) Images

(3) Since Images , f (x ) is discontinuous at x = 3.