AP Calculus AB Unit 3: Differentiation: Composite, Implicit, and Inverse Functions Notes

Introduction

Unit 3 extends differentiation beyond basic formulas. The chain rule follows change through nested functions, while implicit differentiation extracts local slopes from relations that are not solved for y.

Inverse derivatives exchange the roles of input and output. Their slope rule is reciprocal only after the corresponding original input has been found, and it requires a nonzero original slope.

The final work is structural: identify the outer operation, coordinate all needed rules, and interpret higher derivatives as successive rates with changing units.

Current exam guidance: May 2027 format: 42 multiple-choice questions and 6 free-response questions. Calculator availability changes by part; the reader marks calculator decisions separately from mathematical justification.

3.1

The Chain Rule

How do you differentiate a function built inside another function?

Short answer

Differentiate the outside at the unchanged inside, then multiply by the derivative of the inside.

Know firstFunction composition · Basic derivative rules · Product notation

Follow how a change travels through the layers

Suppose u=g(x) and y=f(u). A small change in x first produces about g′(x) times as much change in u. That change then produces about f′(u) times as much change in y. Multiplying those local scale factors gives dy/dx=f′(g(x))g′(x). The chain rule is therefore a rate-conversion rule, not a pattern added for convenience.

For y=(x²+1)³, the outside function is u³ and the inside is x²+1. The outside derivative is 3u²; returning the inside gives 3(x²+1)². The inside changes at rate 2x, so the complete derivative is 6x(x²+1)². Leaving out 2x would pretend the inner input changes one-for-one with x.

Deep compositions are handled one layer at a time. Preserve each inner expression until its layer is differentiated, and multiply every derivative factor. A quick graph check should confirm the derivative sign: here the original falls for x<0, pauses at zero, and rises for x>0.

Formula

Formula and conditions

\frac{d}{dx}f(g(x))=f'(g(x))g'(x)

The derivative of a composition multiplies the local rate of each layer.

Use when: The inner function is differentiable at x and the outer function is differentiable at the corresponding inner value.
Do not assume: Differentiating the outside alone treats g(x) as though its rate were always 1.

The chain factor changes the slope

The composite (x²+1)³ has derivative 6x(x²+1)². The derivative is zero at x=0 because the inner function x²+1 pauses there, even though the outer cube function is changing at input 1.

The interactive version of this figure is part of the full study guide.

The chain factor changes the slope: values

Case	Result
$x$	$f(x)$ · f'(x)
$-1$	$8 · -24$
0	1 · 0
1	8 · 24

Mark the layers before differentiating

Box the innermost expression.
Differentiate the outer form while keeping the inner expression intact.
Multiply by the derivative of the inner expression.
Repeat for any additional nested layer, then check domain and sign.

Keep these three ideas

A composition passes change through successive layers.
Every changing layer contributes a factor.
Keep the original inner expression inside the outer derivative.

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

You just read the first topic free. Your free trial opens the rest of this unit, every AP Calculus AB unit, and all 19 AP courses.

3.1 The Chain RuleRead

3.2 Implicit Differentiation

3.3 Differentiating Inverse Functions

3.4 Differentiating Inverse Trigonometric Functions

3.5 Selecting Procedures for Calculating Derivatives

3.6 Calculating Higher-Order Derivatives

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