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AP Calculus BC, AP exams, 4–7%

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

Unit 3 covers the chain rule, implicit differentiation, and derivatives of inverse functions including the inverse trigonometric functions. Higher-order derivatives and the judgment of which technique to apply round out the unit. The chain rule in particular is a prerequisite for parametric and polar differentiation in Unit 9.

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions at a glance

Position
3 of 10 units
Published weighting
4–7% 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 BC. 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 mixed rules.
Implicit differentiation
Differentiating relations and solving for the derivative, including tangent lines to implicit curves.
Derivatives of inverse functions
Using the inverse relationship to find derivative values without an explicit inverse formula.
Inverse trigonometric derivatives
Arcsine, arccosine, and arctangent derivatives in composite expressions.
Higher-order derivatives
Second and higher derivatives and their use in concavity and motion analysis.
Selecting differentiation procedures
Choosing the correct rule or combination for an unfamiliar expression.

Study tips

How to practice Chain, Implicit, Inverse.

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

  • Do mixed drills where the required rule is not announced, since procedure selection is the tested skill.
  • Practice implicit second derivatives, which appear in BC more often than students expect.
  • Connect the chain rule to Unit 9 explicitly; parametric derivatives are chain-rule applications.

Where this fits

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions is 3 of 10 units in the AP Calculus BC guide, and it carries a published weighting of 4–7% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus BC guideOfficial exam information opens in a new tab

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.

  1. Question 1Foundational

    What is d/dx of e^(3x^2)?

    1. A6x e^(3x^2)
    2. Be^(3x^2)
    3. C6x e^(6x)
    4. D3x^2 e^(3x^2 - 1)
    Show answer

    Answer: A. 6x e^(3x^2)

    The derivative of e^u is e^u times u', and with u = 3x^2 the inside derivative is 6x, giving 6x e^(3x^2). Leaving the answer as e^(3x^2) forgets the chain rule factor, and reducing the exponent treats the exponential as a power function.

    Key insight: An exponential never loses its exponent when differentiated; it gains the inside derivative as a factor.

  2. Question 2Foundational

    For x > 0, what is d/dx of ln(5x)?

    1. A1/x
    2. B5/x
    3. C1/(5x)
    4. D5 ln(x)
    Show answer

    Answer: A. 1/x

    The chain rule gives (1/(5x)) times 5, which simplifies to 1/x. Equivalently, ln(5x) = ln(5) + ln(x), and the constant ln(5) differentiates to 0. Stopping at 1/(5x) forgets the inside derivative, and 5/x multiplies by 5 twice.

    Key insight: A constant multiple inside a logarithm becomes an additive constant, so it disappears on differentiation.

  3. Question 3Intermediate

    The curve x^2 + xy + y^2 = 7 passes through the point (1, 2). What is dy/dx at that point?

    1. A-4/5
    2. B4/5
    3. C-5/4
    4. D-1/2
    Show answer

    Answer: A. -4/5

    Differentiating implicitly gives 2x + y + x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = -(2x + y)/(x + 2y). At (1, 2) that is -(2 + 2)/(1 + 4) = -4/5. Writing the quotient upside down at the last step produces the inverted -5/4.

    Key insight: The xy term needs the product rule, contributing both y and x(dy/dx).

  4. Question 4Intermediate

    If y = sin(2x), what is the second derivative d^2y/dx^2?

    1. A-4sin(2x)
    2. B-2sin(2x)
    3. C4sin(2x)
    4. D-4cos(2x)
    Show answer

    Answer: A. -4sin(2x)

    The first derivative is 2cos(2x), and differentiating again gives 2 times -sin(2x) times 2 = -4sin(2x). Answering -2sin(2x) applies the chain rule only once, and -4cos(2x) differentiates the wrong function on the second pass.

    Key insight: Each differentiation of sin(kx) or cos(kx) contributes another factor of k.

  5. Question 5Advanced

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

    1. A3/√(1 - 9x^2)
    2. B1/√(1 - 9x^2)
    3. C3/√(1 - 3x^2)
    4. D3/(1 + 9x^2)
    Show answer

    Answer: A. 3/√(1 - 9x^2)

    The derivative of arcsin(u) is u'/√(1 - u^2). With u = 3x, u' = 3 and u^2 = 9x^2, giving 3/√(1 - 9x^2). The choice with 1 + 9x^2 in the denominator belongs to arctan, and 1 - 3x^2 forgets to square the coefficient.

    Key insight: Arcsine gives a square root denominator; arctangent gives a denominator with no radical.

Continue through AP Calculus BC

Practice Chain, Implicit, Inverse deliberately.

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