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

Executive Summary

Unit 3 extends differentiation to composites (chain rule), implicitly defined curves, and inverse functions — the tools that unlock most AP exam problems.

The chain rule

If $y = f(g(x))$, then $y' = f'(g(x)) \cdot g'(x)$ — "derivative of the outside, evaluated at the inside, times the derivative of the inside." Chain repeatedly for nested functions: $\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$.

Implicit differentiation

When $y$ is not isolated, differentiate both sides with respect to $x$, treating $y$ as a function: every $y$ term gets a $\frac{dy}{dx}$ factor. Then solve for $\frac{dy}{dx}$. Second derivatives implicitly follow the same pattern.

Inverse functions

If $g = f^{-1}$, then $(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$. For inverse trig: $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$.

Higher-order derivatives

$f''$ is the derivative of $f'$ — the rate of change of the rate of change (acceleration). Notation: $f''$, $\frac{d^2y}{dx^2}$.

Exam traps

Forgetting the chain rule's inner derivative is the #1 error. In implicit differentiation, differentiate EVERY term, including $y$-terms and constants (constants vanish). $(f^{-1})'(a)$ is evaluated at $a$, not at $f^{-1}(a)$ — plug $f^{-1}(a)$ into $f'$ first.