Chain Rule

d/dx[f(g(x))] = f'(g(x)) · g'(x) — outer, then inner.

The idea

Theorem (Chain rule).

Let $g$ be differentiable at $x$ and let $f$ be differentiable at $g(x)$. Then the composition $f(g(x))$ is differentiable at $x$, and $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x).$

Differentiate the outer function with the inner left untouched inside it, then multiply by the derivative of the inner.

The product appears because rates compound along a chain. Write $u = g(x)$ and $y = f(u)$, so that $x$ drives $u$ and $u$ drives $y$. Over a short stretch $u$ changes $g'(x)$ times as fast as $x$, and $y$ changes $f'(u)$ times as fast as $u$, so $y$ changes $f'(u) \cdot g'(x)$ times as fast as $x$.

Note where each derivative is evaluated: $g'$ at $x$, but $f'$ at $g(x)$, because $g(x)$ is the input $f$ receives.

Ways to work on it

• Walkthrough. We differentiate a composite power function together, step by step.
• Proof. Derive the chain rule from the limit definition by multiplying and dividing by the inner change.
• Practice. Two steps. Random (ax + b)^n with small integers.
• Hardest. One step. Inner is quadratic: (ax^2 + b)^n.

Not sure where to start? Take the ten-question placement test.