AP Calculus AB Practice Quiz: Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant 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) = x cot(x) with respect to x?

A)cot(x) - x csc²(x)B)-csc²(x)C)cot(x) + x csc²(x)D)1 - csc²(x)

All Questions (7)

What is the derivative of f(x) = x cot(x) with respect to x?

A) cot(x) - x csc²(x)

B) -csc²(x)

C) cot(x) + x csc²(x)

D) 1 - csc²(x)

Correct Answer: A

This question requires the product rule, (uv)' = u'v + uv'. Let u = x and v = cot(x). Then u' = 1 and v' = -csc²(x). The derivative is f'(x) = u'v + uv' = (1)(cot(x)) + (x)(-csc²(x)), which simplifies to cot(x) - x csc²(x).

Find dy/dx for the function y = sec(x) / (1 + tan(x)).

A) sec(x) / (1 + tan(x))²

B) (sec(x)tan(x) + sec³(x) - sec(x)) / (1 + tan(x))²

C) sec(x)tan(x) / sec²(x)

D) ((1 + tan(x))sec(x)tan(x) - sec³(x)) / (1 + tan(x))²

Correct Answer: D

This requires the quotient rule, (u/v)' = (vu' - uv') / v². Let u = sec(x) and v = 1 + tan(x). Then u' = sec(x)tan(x) and v' = sec²(x). The derivative is dy/dx = ((1 + tan(x))(sec(x)tan(x)) - (sec(x))(sec²(x))) / (1 + tan(x))², which simplifies to ((1 + tan(x))sec(x)tan(x) - sec³(x)) / (1 + tan(x))².

A function is defined as f(x) = tan(x) cos(x). Which of the following is f'(x)?

A) sec²(x)cos(x) - tan(x)sin(x)

B) cos(x)

C) -sin(x)

D) sec(x)

Correct Answer: B

Based on trigonometric identities, the function can be simplified before differentiation. f(x) = tan(x)cos(x) = (sin(x)/cos(x))cos(x) = sin(x). The derivative of sin(x) is cos(x). While using the product rule is possible, it is more complex and would yield the same result after simplification.

If f(x) = sec(x), what is f'(x) when derived using the quotient rule and the identity sec(x) = 1/cos(x)?

A) tan²(x)

B) -cot(x)csc(x)

C) sec(x)tan(x)

D) sin(x)

Correct Answer: C

By rewriting sec(x) as 1/cos(x), the quotient rule can be applied. Let u=1 and v=cos(x). Then u'=0 and v'=-sin(x). The derivative is (v*u' - u*v')/v² = (cos(x)*0 - 1*(-sin(x)))/cos²(x) = sin(x)/cos²(x). This expression can be rearranged using identities to (1/cos(x)) * (sin(x)/cos(x)), which simplifies to sec(x)tan(x).

Find the derivative of the function y = x csc(x) cot(x).

A) csc(x)cot(x) - x csc(x)cot²(x) - x csc³(x)

B) -x csc(x)cot²(x) - x csc³(x)

C) csc(x)cot(x) - x csc²(x)cot(x)

D) -csc(x)cot(x)

Correct Answer: A

This problem requires applying the product rule for three functions, (fgh)' = f'gh + fg'h + fgh'. Let f(x)=x, g(x)=csc(x), and h(x)=cot(x). The derivatives are f'=1, g'=-csc(x)cot(x), and h'=-csc²(x). The derivative is (1)(csc(x)cot(x)) + (x)(-csc(x)cot(x))(cot(x)) + (x)(csc(x))(-csc²(x)), which simplifies to csc(x)cot(x) - x csc(x)cot²(x) - x csc³(x).

Which expression is equivalent to the derivative of f(x) = (cot(x) + 1) / csc(x)?

A) cos(x) - sin(x)

B) -sin(x) - cos(x)

C) sin(x) - cos(x)

D) csc²(x) + cot(x)csc(x)

Correct Answer: A

The most efficient way to solve this is to first simplify the function using trigonometric identities. f(x) = cot(x)/csc(x) + 1/csc(x) = (cos(x)/sin(x))/(1/sin(x)) + sin(x) = cos(x) + sin(x). Then, taking the derivative is straightforward: f'(x) = -sin(x) + cos(x), which is equivalent to cos(x) - sin(x).

Let y = csc(x) - cot(x). Find dy/dx.

A) -csc(x)cot(x) + csc²(x)

B) csc(x)(cot(x) - csc(x))

C) -csc(x)cot(x) - csc²(x)

D) cot(x)(csc(x) - cot(x))

Correct Answer: A

The derivative is found by differentiating each term separately. The derivative of csc(x) is -csc(x)cot(x), and the derivative of cot(x) is -csc²(x). Therefore, dy/dx = -csc(x)cot(x) - (-csc²(x)), which simplifies to -csc(x)cot(x) + csc²(x). This can also be written in factored form as csc(x)(csc(x) - cot(x)).