AP Calculus AB 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

What 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)

What 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 an inverse trigonometric function. The derivative of arctan(x) is a standard result: d/dx(arctan(x)) = 1 / (1 + x^2).

If y = arcsin(5x), what is dy/dx?

A) 5 / sqrt(1 - 25x^2)

B) 1 / sqrt(1 - 25x^2)

C) 5 / (1 + 25x^2)

D) -5 / sqrt(1 - 25x^2)

Correct Answer: A

This problem requires using the chain rule with the formula for the derivative of arcsin(u). Let u = 5x, so u' = 5. The derivative formula is d/dx(arcsin(u)) = u' / sqrt(1 - u^2). Substituting u and u' gives 5 / sqrt(1 - (5x)^2), which simplifies to 5 / sqrt(1 - 25x^2).

Find the derivative of g(x) = arccos(x^3).

A) 3x^2 / sqrt(1 - x^6)

B) -3x^2 / sqrt(1 - x^6)

C) -1 / sqrt(1 - x^6)

D) -3x^2 / (1 + x^6)

Correct Answer: B

This problem applies the chain rule to the derivative of arccos(u). Let u = x^3, which means u' = 3x^2. The formula for the derivative is d/dx(arccos(u)) = -u' / sqrt(1 - u^2). Substituting the expressions for u and u' yields -3x^2 / sqrt(1 - (x^3)^2), which simplifies to -3x^2 / sqrt(1 - x^6).

What is the derivative of h(x) = x * arctan(x)?

A) 1 / (1 + x^2)

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

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

D) x / (1 + x^2)

Correct Answer: B

This question requires the use of the product rule, (fg)' = f'g + fg', in combination with the derivative of an inverse trigonometric function. Let f(x) = x and g(x) = arctan(x). Then f'(x) = 1 and g'(x) = 1 / (1 + x^2). Applying the product rule gives (1 * arctan(x)) + (x * (1 / (1 + x^2))), which simplifies to arctan(x) + x / (1 + x^2).

Let f(x) = arcsec(2x^2). What is f'(x)?

A) 4x / (|2x^2| * sqrt(4x^4 - 1))

B) 1 / (|2x^2| * sqrt(4x^4 - 1))

C) 4x / (1 + 4x^4)

D) -4x / (|2x^2| * sqrt(4x^4 - 1))

Correct Answer: A

This problem uses the chain rule with the derivative of arcsec(u). The formula is d/dx(arcsec(u)) = u' / (|u| * sqrt(u^2 - 1)). Here, u = 2x^2, so u' = 4x. Substituting these into the formula gives 4x / (|2x^2| * sqrt((2x^2)^2 - 1)), which simplifies to 4x / (|2x^2| * sqrt(4x^4 - 1)).

Find dy/dx for y = arccot(e^x).

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

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

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

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

Correct Answer: C

This question requires applying the chain rule. The formula for the derivative of arccot(u) is -u' / (1 + u^2). In this case, u = e^x, so u' = e^x. Applying the chain rule, we get -(e^x) / (1 + (e^x)^2), which simplifies to -e^x / (1 + e^(2x)).

If f(x) = arcsin(cos(x)), what is f'(x) for 0 < x < π?

A) 1

B) -1

C) -tan(x)

D) sin(x) / |sin(x)|

Correct Answer: B

This is a complex chain rule problem. Let u = cos(x), so u' = -sin(x). The derivative of arcsin(u) is u' / sqrt(1 - u^2). Substituting gives f'(x) = -sin(x) / sqrt(1 - cos^2(x)). Using the identity sin^2(x) + cos^2(x) = 1, the denominator becomes sqrt(sin^2(x)) = |sin(x)|. Thus, f'(x) = -sin(x) / |sin(x)|. For the interval 0 < x < π, sin(x) is positive, so |sin(x)| = sin(x). Therefore, f'(x) = -sin(x) / sin(x) = -1.