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Solutions to Differentiation Practice Problems for AP Calculus

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By — McGraw-Hill Professional
Updated on Mar 10, 2014

Practice problems for these solutions can be found at: Differentiation Practice Problems for AP Calculus

Part A The use of a calculator is not allowed.

1. Applying the power rule, =30x4 – 1.
2. Rewrite f (x)= as f (x) = x–1 + x–2/3. Differentiate, f'(x)= –x–2 x–5/3 = –
3. Rewrite
4. Applying the quotient rule,
5. Applying the product rule, u =(3x – 2)5 and v =(x 2 – 1), and then the chain rule,
6. 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)

7. Let u =2x – 1,
8. =10[–csc2(2x – 1)](2)

= – 20 csc2(2x – 1).

9. Using the product rule,

=(3[sec(3x )])+[sec(3x ) tan(3x )](3)[3x ]

=3 sec(3x )+9x sec(3x ) tan(3x )

=3 sec(3x )[1+3x tan(3x )]

10. Using the chain rule, let u = sin(x2 – 4).
11. =10(–sin[sin(x2 – 4)])[cos(x2 – 4)](2x )

= – 20x cos(x2 – 4) sin[sin(x2 – 4)]

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

13. Since 3e 5 is a constant, thus its derivative is 0.
14. =0+(4)(e x )+(e x )(4x )

=4e x +4xe x =4e x (1+x )

15. Part B Calculators are allowed.

16. Using implicit differentiation, differentiate each term with respect to x.
17. 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.
18. 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

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

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

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.

19. You can use the difference quotient to approximate f' (a).
20. Or, you can use the symmetric difference quotient to approximate f' (a).

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

21. Entery1=x 5 +3x – 8. The graph ofy1 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 ).

22. Thus the slope of the tangent toy= ln x . At x =e,y= ln x = ln e = 1, which means the point (e, 1) is on the curve ofy= ln x. Therefore, an equation of the tangent is y – 1= at x =e is (x – e) ory= . [See Figure 6.11-1.]

23. =(2)(sin x )+(cos x )(2x )= 2 sin x +2x cos x
24. Entery1=(x – 1)2/3 +2 in your calculator. The graph ofy1 forms a cusp at x =1. Therefore, f is not differentiable at x =1.
25. Differentiate with respect to x:
1. The expression 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.
2. Therefore the slope is 0, i.e., = 0. An alternate method is to expand sin(π/2+h) as sin(π/2) cos h + cos(π/2) sin h.

3. Using the chain rule, let u =(π – x ).
4. Then, f' (x )=2 cos(π – x )[–sin(π – x )](–1)

=2 cos(π – x ) sin(π – x )

f'(0)=2 cos π sin π =0.

5. Since the degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator, the limit is 0.
6. You can use the difference quotient to approximate f' (a).
7. Or, you can use the symmetric difference quotient to approximate f' (a).

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

8. (See Figure 6.12–1.) Checking the three conditions of continuity:
1. f (3) = 3

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