Differentiation - Review the Knowledge You Need to Score High - 5 Steps to a 5 AP Calculus AB & BC

5 Steps to a 5 AP Calculus AB & BC, 2012-2013 Edition (2011)

STEP 4. Review the Knowledge You Need to Score High

Chapter 6. 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 exams, 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 & Quotient Rules, Chain Rule, and Implicit Differentiation. You will also learn to find the derivatives of trigonometric, exponential, logarithmic, and inverse functions, as well as apply L’Hôpital’s Rule.

Image

Key Ideas

Image Definition of the derivative of a function

Image Power rule, product & quotient rules, and chain rule

Image Derivatives of trigonometric, exponential, and logarithmic functions

Image Derivatives of inverse functions

Image Implicit differentiation

Image Higher order derivatives

Image Indeterminate forms and L’Hôpital’s rule

6.1 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

Image

if this limit exists. (Note that f′(x) is read as f prime of x.) Other symbols of the derivative of a function are:

Image

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

Image

(See Figure 6.1-1.)

Image

Figure 6.1-1

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 6.1-26.1-5 on page 79.)

Image

Figure 6.1-2

Image

Figure 6.1-3

Image

Figure 6.1-4

Image

Figure 6.1-5

Example 1

If f(x) = x2 − 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, Image

Image

(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 Image.

The expression Image 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 Image

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

(See Figure 6.1-6.)

Image

Figure 6.1-6

Example 3

If the function f(x) = x2/3 + 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 6.1-7.)

Image

Figure 6.1-7

Example 4

Using a calculator, find the derivative of f(x) = x2 + 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](x2 + 4x, x)|x = 3. The result is 10.

Image

• 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) = nxn−1.

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

Summary of Derivatives of Algebraic Functions

Image

Example 1

If f(x) = 2x3, find (a) f′(x), (b) f′(1) and (c) f′(0).

Note that (a) f′(x) = 6x2, (b) f′(1) = 6(1)2 = 6, and (c) f′(0) = 0.

Example 2

If Image, find Image and Image (which represents Image.

Note that Image and thus, Image and Image does not exist because the expression Image is undefined.

Example 3

Here are several examples of algebraic functions and their derivatives:

Image

Example 4

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

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

The Sum, Difference, Product, and Quotient Rules

If u and v are two differentiable functions, then

Image

Summary of Sum, Difference, Product, and Quotient Rules

Image

Example 1

Find f′(x) if f(x) = x3 − 10x + 5.

Using the sum and difference rules, you can differentiate each term and obtain f′(x) = 3x2 − 10. Or using your calculator, select the d[Differentiate] function and enter d(x3 − 10x + 5, x) and obtain 3x2 − 10.

Example 2

If y = (3x − 5)(x4 + 8x − 1), find Image.

Using the product rule Image, let u = (3x − 5) and v = (x4 + 8x − 1).

Then Image. Or you can use your calculator and enter d((3x − 5)(x4 + 8x − 1), x) and obtain the same result.

Example 3

If Image, find f′(x).

Using the quotient rule Image, let u = 2x − 1 and v = x + 5. Then Image, x/ = −5. 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) = x2 − 3x + 2 at x = 5.

Find the slope of the tangent to the curve at x = 5 by entering d(x2 − 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).

Image

• Remember that Image and Image. 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 Image or Image.

Example 1

If y = (3x − 5)10, find Image.

Using the chain rule, let u = 3x − 5 and thus, y = u10. Then, Image and Image.

Since Image. Or you can use your calculator and enter d((3x − 5)10, x) and obtain the same result.

Example 2

If Image, find f′(x).

Rewrite Image as Image. Using the product rule, Image.

To find Image, use the chain rule and let u = 25 − x2.

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

Example 3

If Image, find Image.

Using the chain rule, let Image. Then Image.

To find Image, use the quotient rule.

Thus, Image. Substituting this quantity back into Image.

An alternate solution is to use the product rule and rewrite Image 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.

6.2 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

Image

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

Example 1

If y = 6x2 + 3 sec x, find Image.

Image.

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 Image which is an equivalent form.

Example 3

Find f′(x) if f(x) = 8 sin(x2).

Using the chain rule, let u = x2. Then f′(x) = [8 cos(x2)][2x] = 16x cos(x2).

Example 4

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

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

Then Image.

Example 5

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

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

Image.

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

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 Image, find Image.

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

Example 8

If Image, find Image.

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

Image

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

Image

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 Image.

Let u = 3x. Then Image.

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

Example 2

Find f′(x) if Image.

Let

Image

Example 3

If y = sec−1(3x2), find Image.

Let u = 3x2. Then Image.

Example 4

If Image, find Image.

Let Image. Then Image.

Rewrite Image as u = x−1. Then Image.

Therefore,

Image

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

Image

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

Example 1

y = e3x + 5xe3 + e3

Image (Note that e3 is a constant.)

Example 2

y = xexx2ex

Using the product rule for both terms, you have

Image

Example 3

y = 3sin x

Let u = sin x. Then, Image.

Example 4

Image

Let u = x3. Then, Image.

Example 5

y = (ln x)5

Let u = ln x. Then, Image.

Example 6

y = ln(x2 + 2x − 3) + ln 5

Let u = x2 + 2x − 3. Then, Image.

(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 Image.

Example 8

y = ln(ln x)

Let u = ln x. Then Image.

Example 9

y = log5(2x + 1)

Let u = 2x + 1. Then Image.

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 = ex 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 = e1 = 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 6.2-1.)

Image

Figure 6.2-1

Image

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

6.3 Implicit Differentiation

Main Concept: Procedure for Implicit Differentiation

Procedure for Implicit Differentiation

Image

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

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

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

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

4: Solve for Image.

Example 1

Find Image if y2 − 7y + x2 − 4x = 10.

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

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

Step 3: Factor out Image.

Step 4: Solve for Image.

Example 2

Given x3 + y3 = 6xy, find Image.

Step 1: Differentiate each term with respect to Image.

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

Step 3: Factor out Image.

Step 4: Solve for Image.

Example 3

Find Image if (x + y)2 − (xy)2 = x5 + y5.

Step 1: Differentiate each term with respect to x:

Image

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

Image

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

Image

Step 3: Factor out Image:

Image

Step 4: Solve for Image.

Example 4

Write an equation of the tangent to the curve x2 + y2 + 19 = 2x + 12y at (4, 3).

The slope of the tangent to the curve at (4, 3) is equivalent to the derivative Image at (4, 3).

Using implicit differentiation, you have:

Image

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

Example 5

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

Image

6.4 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 6.4-1. Find the approximate value of f′(3). (See Figure 6.4-1.)

Image

Figure 6.4-1

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: Using the slope of the line segment joining the points at x = 3 and x = 4.

f(3) = 3 and f(4) = 5

Image

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

f(2) = 2 and f(3) = 3

Image

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

f(2) = 2 and f(4) = 5

Image

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

Thus, f′(3) ≈ 1, 2 or Image 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.

Image

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

Image

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

Image

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 6.4-2.)

Image

Figure 6.4-2

Image

• Remember that the Image because the Image.

6.5 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) = Image. (See Figure 6.5-1.)

Image

Figure 6.5-1

If y = f−1(x) so that x = f(y), then Image.

Example 1

If f(x) = x3 + 2x − 10, find (f−1)′(x).

Step 1: Check if (f−1)′(x) exists. f′(x) = 3x2 + 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 = x3 + 2x − 10.

Step 3: Interchange x and y to obtain the inverse function x = y3 + 2y − 10.

Step 4: Differentiate with respect to Image.

Step 5: Apply formula Image.

Image. Thus, Image.

Example 2

Example 1 could have been done by using implicit differentiation.

Step 1: Let y = f(x), and thus y = x3 + 2x − 10.

Step 2: Interchange x and y to obtain the inverse function x = y3 + 2y − 10.

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

Image

Step 4: Solve for Image.

Image

Example 3

If f(x) = 2x5 + x3 + 1, find (a) f(1) and f′(1) and (b) (f−1)(4) and (f−1)′(4). Enter y1 = 2x5 + x3 + 1. Since y1 is strictly increasing, thus f(x) has an inverse.

(a) f(1) = 2(1)5 + (1)3 + 1 = 4

f′(x) = 10x4 + 3x2

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) = 2x5 + x3 + 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.

Image

Example 4

If f(x) = 5x3 + x + 8, find (f−1)′(8).

Enter y1 = 5x3 + x + 8. Since y1 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).

f′(x) = 15x2 + 1

f′(0) = 1

Therefore, Image.

Image

• 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.

6.6 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

Image

Note that Image or Image.

Example 1

If y = 5x3 + 7x − 10, find the first four derivatives.

Image

Example 2

If Image, find f″(4).

Rewrite: Image and differentiate: Image.

Differentiate again:

Image

Example 3

If y = x cos x, find y″.

Using the product rule, y′ = (1)(cos x) + (x)(−sin x) = cos xx sin x

y″ = −sin x − [(1)(sin x) + (x)(cos x)]

= −sin x − sin xx cos x

= − 2sin xx cos x.

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

6.7 Indeterminate Forms

Image

Main Concept: L’Hôpital’s Rule for Indeterminate Forms

L’Hôpital’s Rule for Indeterminate Forms

Suppose f(x) and g(x) are differentiable, and g′(x) ≠ 0 near c, except possibly at c, and suppose Image and Image so that Image is an indeterminate form. In this case Image. If Image and Image, the same rule applies.

Example 1

Find Image, if it exists.

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

Example 2

Find Image, if it exists.

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

6.8 Rapid Review

1. If Image, find Image.

Answer: Using the chain rule, Image.

2. Evaluate Image.

Answer: The limit is equivalent to Image.

3. Find f′(x) if f(x) = ln(3x).

Answer: Image.

4. Find the approximate value of f′(3). (See Figure 6.8-1.)

Image

Figure 6.8-1

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

5. Find Image if xy = 5x2.

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

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

6. If Image, find Image.

Answer: Rewrite y = 5x−2. Then, Image and Image.

7. Using a calculator, write an equation of the line tangent to the graph f(x) = −2x4 at the point where f′(x) = −1.

Answer: f′(x) = −8x3. Using a calculator, enter [Solve][−8x^3 = −1, x] and obtain Image. Using the calculator Image. Thus, tangent is Image.

Image

8. Image

Answer: Since Image, consider Image.

9. Image

Answer: Since Image, consider Image.

6.9 Practice Problems

Part A—The use of a calculator is not allowed.

Find the derivative of each of the following functions.

1. y = 6x5x + 10

2. Image

3. Image

4. Image

5. f(x) = (3x − 2)5(x2 − 1)

6. Image

7. y = 10 cot(2x − 1)

8. y = 3x sec(3x)

9. y = 10 cos[sin(x2 − 4)]

10. y = 8 cos−1(2x)

11. y = 3e5 + 4xex

12. y = ln(x2 + 3)

Part B—Calculators are allowed.

13. Find Image, if x2 + y3 = 10 − 5xy.

14. The graph of a function f on [1, 5] is shown in Figure 6.9-1. Find the approximate value of f′(4).

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.

Image

Image

Figure 6.9-1

16. If f(x) = x5 + 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 Image at Image.

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 Image.

Image

21. Image

22. Image

23. Image

24. Image

25. Image

6.10 Cumulative Review Problems

(Calculator) indicates that calculators are permitted.

26. Find Image.

27. If f(x) = cos2(πx), find f′(0).

28. Find Image.

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.

Image

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

6.11 Solutions to Practice Problems

Part A—The use of a calculator is not allowed.

1. Applying the power rule, Image.

2. Rewrite Image as f(x) = x−1 + x−2/3. Differentiate:

Image

3. Rewrite

Image

Differentiate:

Image

An alternate method is to differentiate Image directly, using the quotient rule.

4. Applying the quotient rule,

Image

5. Applying the product rule, u = (3x − 2)5 and v = (x2 − 1), and then the chain rule,

f′(x) = [5(3x − 2)4(3)][x2 − 1] + [2x] × [(3x −2)5]

= 15(x2 − 1)(3x − 2)4 + 2x(3x − 2)5

= (3x − 2)4[15(x2 − 1) + 2x(3x − 2)]

= (3x − 2)4[15x2 − 15 + 6x2 − 4x]

= (3x − 2)4(21x2 − 4x − 15).

6. Rewrite Image as Image Applying first the chain rule and then the quotient rule,

Image

Note: Image, if Image which implies Image or Image.

An alternate method of solution is to write Image and use the quotient 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,

Image

8. Using the product rule,

Image

9. Using the chain rule, let u = sin(x2 − 4).

Image

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

Image

11. Since 3e5 is a constant, its derivative is 0.

Image

12. Let

Image

Part B—Calculators are allowed.

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

Image

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: Using the slope of the line segment joining the points at x = 4 and x = 5.

f(5) = 1 and f(4) = 4

Image

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

f(3) = 5 and f(4) = 4

Image

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

f(3) = 5 and f(5) = 1

Image

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 Image to approximate f′(a).

Let h = 1; Image.

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

Let h = 1; Image

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

16. Enter y 1 = x5 + 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) = 5x4 + 3 and f′(0) = 3.

Since Image, thus Image.

17. Image and Image

Thus the slope of the tangent to y = ln x at x = e is Image. 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 Image or Image. (See Figure 6.11-1.)

Image

Figure 6.11-1

18. Image

Image

Image

Or, using a calculator, enter

d(2x − sin(x), x, 2) Image 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:

Image

Image

Thus, the slope of the tangent to the curve at (2, π/3) is Image.

The slope of the normal line to the curve at (2, π/3) is Image

Therefore an equation of the normal line is Image.

21.

Image

22.

Image

23.

Image

24.

Image

25.

Image

6.12 Solutions to Cumulative Review Problems

26. The expression Image is 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., Image

An alternate method is to expand sin(π/2 + h) as

sin(π/2) cos h + cos(π/2) sin h.

Thus,

Image

Image

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

Then, f′(x) = 2 cos(πx)[− sin(πx)](−1)

= 2 cos(πx) sin(πx)

f′(0) = 2 cos π sin π = 0.

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 difference quotient Image to approximate f′(a).

Let h = 1;

Image

Let h = 2;

Image

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

Let h = 1;

Image

Let h = 2;

Image

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

30. (See Figure 6.12–1.) Checking the three conditions of continuity:

Image

Figure 6.12-1

(1) f(3) = 3

(2)

Image

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