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Mathematics and Statistics · Ch 3 — Differentiation

Derivative of Composite Functions — the Chain Rule

3

Derivative of Composite Functions — the Chain Rule

A composite function is a "function of a function", such as (3x2+5)4(3x^2+5)^4 (a power applied to a polynomial) or log⁡(x2+1)\log(x^2+1) (a logarithm applied to a polynomial). Writing the outer function as y=f(u)y = f(u) and the inner as u=g(x)u = g(x), the chain rule states:

dydx=dydu⋅dudx.\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

In words: differentiate the outer function (treating the inner as a single block), then multiply by the derivative of the inner function. The two dudu's "cancel" as a memory aid, though the rule is a genuine theorem, not fraction cancellation.

Common special cases (with uu a function of xx):

ddx(un)=n un−1 dudx,ddx(eu)=eu dudx,ddx(log⁡u)=1u dudx.\frac{d}{dx}\big(u^n\big) = n\,u^{n-1}\,\frac{du}{dx}, \qquad \frac{d}{dx}\big(e^{u}\big) = e^{u}\,\frac{du}{dx}, \qquad \frac{d}{dx}\big(\log u\big) = \frac{1}{u}\,\frac{du}{dx}.

Method — chain rule step by step:

  1. Identify the inner function u=g(x)u = g(x) (usually what sits inside a bracket, root, power, log or trig function).
  2. Differentiate the outer function with respect to uu, leaving uu untouched inside.
  3. Multiply by dudx\dfrac{du}{dx}, the derivative of the inner function.
  4. For deeply nested functions, apply the rule repeatedly, working from the outside in — one factor per layer.

Illustration. For y=(x2+1)3y = (x^2+1)^3: the inner is u=x2+1u = x^2+1 with dudx=2x\dfrac{du}{dx}=2x; the outer is u3u^3 with dydu=3u2\dfrac{dy}{du}=3u^2. So dydx=3(x2+1)2⋅2x=6x(x2+1)2\dfrac{dy}{dx}=3(x^2+1)^2 \cdot 2x = 6x(x^2+1)^2.

Note

Never Forget to Multiply by the Derivative of the Inner Function …

Definition 5Chain rule (composite functions)

If y=f(u)y=f(u) and u=g(x)u=g(x), then dydx=dydu⋅dudx\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx} — differentiate the outer function with the inner treated as one block, then multiply by the derivative of the inner. Special cases: ddxun=nun−1dudx\tfrac{d}{dx}u^n=n u^{n-1}\tfrac{du}{dx}, $\tfrac{d}{dx}e^u=e^u\tfrac{du} …