AP Calculus BC Practice Quiz: Differentiating Inverse Trigonometric Functions

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

Which of the following is the derivative of f(x) = arctan(x) with respect to x?

A)1 / (1 + x^2)B)-1 / (1 + x^2)C)1 / sqrt(1 - x^2)D)-1 / sqrt(1 - x^2)

All Questions (7)

Which of the following is the derivative of f(x) = arctan(x) with respect to x?

A) 1 / (1 + x^2)

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

C) 1 / sqrt(1 - x^2)

D) -1 / sqrt(1 - x^2)

Correct Answer: A

This question tests the basic formula for the derivative of the inverse tangent function. The derivative of arctan(x) is 1 / (1 + x^2). Option C is the derivative of arcsin(x), and Option D is the derivative of arccos(x).

If y = arcsin(3x), then what is dy/dx?

A) 3 / (1 + 9x^2)

B) 1 / sqrt(1 - 9x^2)

C) 3 / sqrt(1 - 9x^2)

D) 3 / sqrt(1 - 3x^2)

Correct Answer: C

This requires the chain rule. Let u = 3x, so du/dx = 3. The derivative of arcsin(u) is 1/sqrt(1 - u^2). Applying the chain rule, dy/dx = (1 / sqrt(1 - (3x)^2)) * 3 = 3 / sqrt(1 - 9x^2).

Let f(x) = arctan(e^x). What is f'(x)?

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

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

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

D) e^x / sqrt(1 - e^(2x))

Correct Answer: B

This problem requires the application of the chain rule. The outer function is arctan(u) and the inner function is u = e^x. The derivative of the outer function is 1/(1+u^2) and the derivative of the inner function is e^x. By the chain rule, f'(x) = (1 / (1 + (e^x)^2)) * e^x = e^x / (1 + e^(2x)).

What is the derivative of g(x) = x * arccos(x)?

A) arccos(x) - x / sqrt(1 - x^2)

B) 1 - 1 / sqrt(1 - x^2)

C) -x / sqrt(1 - x^2)

D) arccos(x) + x / sqrt(1 - x^2)

Correct Answer: A

This problem requires the product rule in combination with the derivative of an inverse trigonometric function. Using the product rule (uv)' = u'v + uv', let u = x and v = arccos(x). Then u' = 1 and v' = -1/sqrt(1 - x^2). Therefore, g'(x) = (1) * arccos(x) + x * (-1 / sqrt(1 - x^2)) = arccos(x) - x / sqrt(1 - x^2).

If f(x) = arcsin(x/2), what is the value of f'(1)?

A) 1 / sqrt(3)

B) 2 / sqrt(3)

C) 1 / (2 * sqrt(3))

D) 1/2

Correct Answer: A

First, find the derivative f'(x) using the chain rule. Let u = x/2, so du/dx = 1/2. The derivative of arcsin(u) is 1/sqrt(1-u^2). So, f'(x) = (1 / sqrt(1 - (x/2)^2)) * (1/2) = 1 / (2 * sqrt(1 - x^2/4)). Now, evaluate at x=1: f'(1) = 1 / (2 * sqrt(1 - 1/4)) = 1 / (2 * sqrt(3/4)) = 1 / (2 * sqrt(3)/2) = 1 / sqrt(3).

Let h(x) = arcsin(x) + arccos(x). For -1 < x < 1, what is h'(x)?

A) 2 / sqrt(1 - x^2)

B) 0

C) pi / 2

D) 2 / (1 + x^2)

Correct Answer: B

There are two ways to solve this. Method 1: Differentiate term by term. The derivative of arcsin(x) is 1/sqrt(1-x^2) and the derivative of arccos(x) is -1/sqrt(1-x^2). Their sum is 0. Method 2: Use the trigonometric identity arcsin(x) + arccos(x) = pi/2. Since h(x) is a constant, its derivative h'(x) is 0.

What is the slope of the line tangent to the graph of y = arctan(sqrt(x)) at x = 3?

A) 1/8

B) 1 / (8 * sqrt(3))

C) 1 / (4 * sqrt(3))

D) 1/4

Correct Answer: B

The slope of the tangent line is the value of the derivative at that point. We must use the chain rule to find dy/dx. Let u = sqrt(x), so du/dx = 1/(2*sqrt(x)). The derivative of arctan(u) is 1/(1+u^2). Thus, dy/dx = (1 / (1 + (sqrt(x))^2)) * (1 / (2*sqrt(x))) = 1 / ((1+x) * 2*sqrt(x)). Now, evaluate at x=3: dy/dx = 1 / ((1+3) * 2*sqrt(3)) = 1 / (4 * 2*sqrt(3)) = 1 / (8 * sqrt(3)).