AP Calculus AB Practice Quiz: The Quotient Rule

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Let f(x) = (2x + 1) / (x - 3). What is f'(x)?

A)7 / (x - 3)^2B)-7 / (x - 3)^2C)(4x - 5) / (x - 3)^2D)-7 / (x - 3)

All Questions (7)

Let f(x) = (2x + 1) / (x - 3). What is f'(x)?

A) 7 / (x - 3)^2

B) -7 / (x - 3)^2

C) (4x - 5) / (x - 3)^2

D) -7 / (x - 3)

Correct Answer: B

Using the quotient rule, d/dx [u/v] = (v*u' - u*v') / v^2. Let u = 2x + 1, so u' = 2. Let v = x - 3, so v' = 1. Then f'(x) = [(x - 3)(2) - (2x + 1)(1)] / (x - 3)^2 = [2x - 6 - 2x - 1] / (x - 3)^2 = -7 / (x - 3)^2.

If y = sin(x) / x^2, then dy/dx =

A) (x cos(x) - 2 sin(x)) / x^3

B) (2 sin(x) - x cos(x)) / x^3

C) (x cos(x) + 2 sin(x)) / x^3

D) cos(x) / 2x

Correct Answer: A

Apply the quotient rule, (g(x)f'(x) - f(x)g'(x)) / [g(x)]^2, where f(x) = sin(x) and g(x) = x^2. Then f'(x) = cos(x) and g'(x) = 2x. The derivative is [x^2 * cos(x) - sin(x) * 2x] / (x^2)^2 = [x^2*cos(x) - 2x*sin(x)] / x^4. Factoring out an x from the numerator and simplifying gives (x cos(x) - 2 sin(x)) / x^3.

Let h(x) = (3x - 2) / (x^2 + 1). What is the value of h'(1)?

A) 1/2

B) 2

C) -1

D) 1

Correct Answer: D

First, find the derivative h'(x) using the quotient rule: h'(x) = [(x^2 + 1)(3) - (3x - 2)(2x)] / (x^2 + 1)^2. This simplifies to h'(x) = [3x^2 + 3 - 6x^2 + 4x] / (x^2 + 1)^2 = (-3x^2 + 4x + 3) / (x^2 + 1)^2. Now, substitute x = 1: h'(1) = (-3(1)^2 + 4(1) + 3) / (1^2 + 1)^2 = (-3 + 4 + 3) / (2)^2 = 4 / 4 = 1.

Let f and g be differentiable functions. If h(x) = f(x) / g(x), which of the following is h'(x)?

A) (f(x)g'(x) - g(x)f'(x)) / [g(x)]^2

B) (g(x)f'(x) - f(x)g'(x)) / [g(x)]^2

C) (g(x)f'(x) + f(x)g'(x)) / [g(x)]^2

D) f'(x) / g'(x)

Correct Answer: B

This is the definition of the quotient rule. The derivative of a quotient f(x)/g(x) is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. This corresponds to the expression (g(x)f'(x) - f(x)g'(x)) / [g(x)]^2.

The functions f and g are differentiable. The table below gives values of the functions and their derivatives at x = 2. | x | f(x) | f'(x) | g(x) | g'(x) | |---|---|---|---|---| | 2 | 3 | -1 | 5 | 4 | If h(x) = f(x) / g(x), what is the value of h'(2)?

A) 17/25

B) 7/25

C) -17/25

D) 7

Correct Answer: C

Using the quotient rule, h'(x) = [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2. To find h'(2), substitute the values from the table: h'(2) = [g(2)f'(2) - f(2)g'(2)] / [g(2)]^2 = [(5)(-1) - (3)(4)] / (5)^2 = [-5 - 12] / 25 = -17/25.

If f(x) = (x^2 - 9) / (x - 3) for x ≠ 3, then f'(x) =

A) 1

B) -1

C) 2x

D) (x^2 - 6x + 9) / (x - 3)^2

Correct Answer: A

While the quotient rule can be used, it is much simpler to first simplify the function. f(x) = (x^2 - 9) / (x - 3) = (x - 3)(x + 3) / (x - 3) = x + 3 for x ≠ 3. The derivative of f(x) = x + 3 is f'(x) = 1. Applying the quotient rule without simplifying would also yield 1 after algebraic manipulation.

What is the derivative of y = e^x / (x^2 + 1)?

A) (e^x (x^2 - 2x + 1)) / (x^2 + 1)^2

B) (e^x (x^2 + 2x + 1)) / (x^2 + 1)^2

C) (e^x (2x - x^2 - 1)) / (x^2 + 1)^2

D) e^x / 2x

Correct Answer: A

Apply the quotient rule, (v*u' - u*v') / v^2, with u = e^x and v = x^2 + 1. We have u' = e^x and v' = 2x. The derivative is [(x^2 + 1)(e^x) - (e^x)(2x)] / (x^2 + 1)^2. Factoring out e^x from the numerator gives [e^x (x^2 + 1 - 2x)] / (x^2 + 1)^2, which simplifies to (e^x (x^2 - 2x + 1)) / (x^2 + 1)^2. This can also be written as (e^x (x - 1)^2) / (x^2 + 1)^2.