CALCULUS AB DERITVATIVE PRACTICE PROBLEMS

Problem 1 :

If y = x sin x, then dy/dx

A)  sin x + cos x       B)  sin x + x cos x      C)  sin x - x cos x

D)  x(sin x + cos x)       E)  x(sin x - cos x)

Solution :

y = x sin x

To find the derivative, we use the product rule.

u = x and v = sin x

u' = 1 and v' = cos x

dy/dx = x cos x + sin x (1)

dy/dx = x cos x + sin x

So, option B is correct.

Problem 2 :

Let f be the function given by f(x) = 300x - x3. On which intervals is the function f increasing ?

A)  (-∞, -10] and [10, ∞)      B)  [-10, 10]     C)  [0, 10] only 

D)  [0, 10√3] only     E)  [0, ∞)

Solution :

f(x) = 300x - x3

f'(x) = 300(1) - 3x2

f'(x) = 3(100 - x2)

f'(x) = 0

3(100 - x2) = 0

100 - x2 = 0 or (10 + x) (10 - x) = 0

x = -10 and 10

calculus-ab-practice-q2


(-∞, -10)

(-10, 10)

(10, ∞)

f'(x)

-

+

-

Increasing/decreasing

Decreasing

Increasing

Decreasing

Increasing --> Positive slope [f'(x) > 0]

At [-10, 10], the function f(x) is having positive slope. So, option B is correct.

Problem 3 :

If f(x) = 7x - 3 + ln x, then f'(1) = ?

A)  4     B)  5      C)  6    D)  7     E)  8

Solution :

f(x) = 7x - 3 + ln x

f'(x) = 7(1) - 0 + 1/x

f'(x) = 7 + 1/x

f'(1) = 7 + 1/1 ==> 8

So, option E will be correct.

Problem 4 :

calculus-ab-practice-q4.png
A) limx2 f(x) exists B) limx3 f(x) existsC) limx4 f(x) exists D) limx5 f(x) existsE) the function is continuous at x = 3

Solution :

A) limx2 f(x) exists limx2-limx2+ B) limx3 f(x) exists limx3-limx3+C) limx4 f(x) exists limx4-limx4+

In option C, the limit does not exists. So, option C is correct.

Problem 5 :

If y = (x3 - cos x)5, then find y'

A)  5(x3 - cos x)4            B)  5(3x2 + sin x )

C)  5(3x2 + sin x)         D)  5(3x2 + sin x)4 (6x + cos x)

E)  5(x3 - cos x)4 (3x2 + sin x)

Solution :

y = (x3 - cos x)5

Using chain rule, we find the derivative.

y' = 5(x3 - cos x)4 (3x2 - (-sin x))

y' = 5(x3 - cos x)4 (3x2 + sin x)

So, option E is correct.

Problem 6 :

If

f(x) = √(x2-4) and g(x) = 3x-2

then find the derivative of f((g(x)) at x = 3

A)  7/√5     B)  14/√5      C)  18/√5    D)  15/√21     E)  30/√21

Solution :

f(x) = √(x- 4) and g(x) = 3x - 2

f(g(x)) = √((3x - 2)- 4)

√[(3x)2 - 2(3x)(2) + 2- 4]

√(9x2 - 12x)

[f(g(x))]' = [1/2√(9x2 - 12x)] (18x - 12)

= [1/2√(9x2 - 12x)] (18x - 12)

[f(g(x))]'  at x = 3 :

= [1/2√(81 - 36)] (54 - 12)

= [21/√45]

= 21/3√5

= 7/√5

So, option A is correct.

Problem 7 :

The function f is defined by f(x) = x/(x + 2). What points (x, y) on the graph of f have the property that the line tangent to f at (x, y) has slope 1/2 ?

A)  (0, 0) only       B)  (1/2, 1/5) only     C)  (0, 0) and (-4, 2)

D)  (0, 0) and (4, 2/3)    E) There are no such points

Solution :

f(x)=xx+2f'(x)=(x+2) 1-x(1+0)(x+2)2=x+2-x(x+2)2f'(x)=2(x+2)2Given slope=1212=2(x+2)2(x+2)2 = 4x+ 2 = 4x + 2= ±2x+ 2=2 amd x + 2=-2x=0 and x=-4

When x = 0

y = 0/2

y = 0

When x = -4

y = -4/(-2)

y = 2

So, the points are (0, 0) and (-4, 2). Then option C is correct.

Problem 8 :

Let f(x) = (2x+1)3 and let g be the inverse function of f. Given that f(0) = 1, what is the value of g'(1) ?

A)  -2/27      B)  1/54    C)  1/27   D)  1/6    E)  6

Solution :

Given f(0) = 1

0 = f-1(1)

g is the function inverse of f, g(1) = 0.

f(x) = (2x+1)3

f'(x) = 3(2x+1)2 (2)

f'(x) = 6(2x+1)2

f'(0) = 6

f'(f-1(1)) = 6

f'(g(1)) = 6

g(1) = 1/f'(g(1))

g(1) = 1/6

So, option D is correct.

Problem 9 :

Let f be the function defined by f(x) = ln x/x. What is absolute maximum value of f ?

A)  1    B)  1/e      C)  0       D) -e

E) f does not have an absolute maximum value

Solution :

f(x) = ln x/x

f'(x) = [x(1/x) - ln x(1)] / x2

f'(x) = [1 - ln x] / x2

f'(x) = 0

1 - ln x = 0

ln x = 1

x = e

Since the given function involves ln, we are allowed to use positive values only.

(0, e), (e, ∞)

x = 1 ∈ (0, e) ==>  f'(1) = [1 - ln 1] / 1 ==> 1 > 0

x = 3  (e, ∞) ==> f'(4) = [1 - ln 4] / 4 ==> - < 0

Drawing the sign diagram, we get

calculus-ab-practice-q9.png

There should be absolute maximum at x = e

Maximum Value :

f(e) = ln e/e

f(e) = 1/e

So, the maximum value is 1/e.

Problem 10 :

Let g be the function given by g(x) = x2 ekx, where k is a constant. For what value of k does g have critical point at x = 2/3 ?

A)  -3    B)  -3/2    C)  -1/3      D)  0     E) There is no such k.

Solution :

g(x) = x2 ekx

u = x2 and v =  ekx

u' = 2x and v' = kekx

g'(x) = x2(kekx) + ekx (2x)

g'(x) = xekx (kx + 2]

g'(x) = 0 at x = 2/3

(2/3)ek(2/3) [k(2/3) + 2] = 0

ek(2/3) = 0

k(2/3) = ln0

2k/3 = 1

k = 3/2

2k/3 + 2 = 0

2k/3 = -2

k = -2(3/2)

k = -3

So, option A is correct.

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