AP Calculus AB, AP exams, 9–13%

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

Unit 3 extends differentiation to the cases that matter most in applications. The chain rule enables everything that follows, implicit differentiation handles relations that are not solved for a variable, and the inverse function rule covers inverse trigonometric and logarithmic derivatives. Higher-order derivatives are introduced here as well.

Position: 3 of 8 units
Published weighting: 9–13% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Calculus AB. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

The chain rule: Differentiating composite functions, including nested compositions and combinations with other rules.
Implicit differentiation: Differentiating relations term by term and solving for the derivative, including tangent lines to curves.
Derivatives of inverse functions: Using the inverse function relationship to find a derivative without an explicit formula.
Inverse trigonometric derivatives: Derivatives of arcsine, arccosine, and arctangent and their use in composite expressions.
Higher-order derivatives: Second and higher derivatives, their notation, and their interpretation in motion problems.
Selecting procedures for derivatives: Recognizing which rule or combination of rules a given expression requires before differentiating.

Study tips

How to practice Chain, Implicit, Inverse.

Section-specific guidance that follows from how this material is usually tested.

Identify the outer and inner functions explicitly before applying the chain rule until the recognition is automatic.
Practice implicit differentiation on curves where the tangent is vertical or horizontal, a common exam prompt.
Do mixed derivative drills where the rule is not announced; procedure selection is itself a tested skill.

Where this fits

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions is 3 of 8 units in the AP Calculus AB guide, and it carries a published weighting of 9–13% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus AB guide Official exam information

Sample Chain, Implicit, Inverse questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
If f(x) = (3x^2 + 1)^5, what is f'(x)?
1. A5(3x^2 + 1)^4
2. B30x(3x^2 + 1)^4
3. C6x(3x^2 + 1)^4
4. D30x(3x^2 + 1)^5
Show answer
Answer: B. 30x(3x^2 + 1)^4
The chain rule gives 5(3x^2 + 1)^4 times the derivative of the inside, 6x, so f'(x) = 30x(3x^2 + 1)^4. Choosing 5(3x^2 + 1)^4 omits the inside derivative, the most common chain rule error, and the last choice fails to reduce the exponent.
Key insight: Differentiate the outside, keep the inside, then multiply by the inside derivative.
Question 2Foundational
What is d/dx of sin(4x)?
1. A4cos(4x)
2. Bcos(4x)
3. C-4cos(4x)
4. D4sin(4x)
Show answer
Answer: A. 4cos(4x)
The derivative of sin(u) is cos(u) times du/dx, and here u = 4x so du/dx = 4, giving 4cos(4x). Answering cos(4x) forgets the inside derivative, and the negative sign belongs to the derivative of cosine, not sine.
Key insight: Sine differentiates to cosine with no sign change; the minus sign appears when differentiating cosine.
Question 3Intermediate
The curve x^2 + y^2 = 25 passes through the point (3, 4). What is the slope of the line tangent to the curve at that point?
1. A-3/4
2. B3/4
3. C-4/3
4. D4/3
Show answer
Answer: A. -3/4
Differentiating implicitly gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y. At (3, 4) that is -3/4. The value -4/3 comes from writing dy/dx = -y/x instead of -x/y, and a positive 3/4 drops the sign that implicit differentiation produces.
Key insight: Implicit differentiation gives a slope formula in both x and y, so a point is needed to evaluate it.
Question 4Intermediate
Let f(x) = x^3 + x + 1, so that f(1) = 3, and let g be the inverse function of f. What is g'(3)?
1. A1/4
2. B1/3
3. C3
4. D4
Show answer
Answer: A. 1/4
The inverse function rule says g'(b) = 1/f'(a) where f(a) = b. Here a = 1 and f'(x) = 3x^2 + 1, so f'(1) = 4 and g'(3) = 1/4. Answering 4 reports f'(1) itself instead of its reciprocal.
Key insight: The derivative of an inverse is the reciprocal of the original derivative at the matching point.
Question 5Advanced
If y = arctan(x^2), what is dy/dx?
1. A2x/(1 + x^4)
2. B2x/(1 + x^2)
3. C1/(1 + x^4)
4. D2x/(1 + 2x^2)
Show answer
Answer: A. 2x/(1 + x^4)
The derivative of arctan(u) is u'/(1 + u^2). With u = x^2, u^2 = x^4 and u' = 2x, so dy/dx = 2x/(1 + x^4). Writing 1 + x^2 in the denominator forgets to square the inside function, and dropping the 2x omits the chain rule factor.
Key insight: In arctan(u) the denominator is 1 + u^2, so squaring the inside function comes before anything else.

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Unit 3: Differentiation: Composite, Implicit, and Inverse Functions