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

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 :

Solution :

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 )⁴ 

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) = √(x² - 4) and g(x) = 3x - 2

f(g(x)) = √((3x - 2)² - 4)

= √[(3x)² - 2(3x)(2) + 2² - 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 :

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] / 1²  ==> 1 > 0

x = 3 ∈ (e, ∞) ==> f'(4) = [1 - ln 4] / 4²  ==> - < 0

Drawing the sign diagram, we get

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 e^kx, 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 e^kx

u = x² and v =  e^kx

u' = 2x and v' = ke^kx

g'(x) = x²(ke^kx) + e^kx (2x)

g'(x) = xe^kx (kx + 2]

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

(2/3)e^k(2/3) [k(2/3) + 2] = 0

So, option A is correct.