Chain rule derivative: never get lost

The formula

If $u(x) = f\bigl(g(x)\bigr)$, then:

$u'(x) = f'\bigl(g(x)\bigr) \times g'(x)$

In English: derivative of the outer function evaluated at the inner function, times the derivative of the inner.

The mnemonic that always works

When you see $f(g(x))$, switch to « onion » mode: two layers to peel.

Differentiate the outer, keep the inner, multiply by the derivative of the inner.

A step-by-step example

Differentiate $u(x) = (3x + 2)^5$.

Three quick examples

$\bigl(\sin(2x)\bigr)' = \cos(2x) \times 2 = 2\cos(2x)$ (outer: sin. Inner: 2x, derivative = 2.)

$\bigl(e^{x^2}\bigr)' = e^{x^2} \times 2x$ (outer: exp. Inner: x², derivative = 2x.)

$\bigl(\sqrt{x^2 + 1}\bigr)' = \dfrac{1}{2\sqrt{x^2 + 1}} \times 2x = \dfrac{x}{\sqrt{x^2 + 1}}$ (outer: sqrt. Inner: x² + 1, derivative = 2x.)

The classic mistake

Forgetting to multiply by the inner’s derivative. $\bigl(\sin(2x)\bigr)' = \cos(2x)$ is wrong — missing the factor 2.

Quick test: if the inner isn’t just $x$, you need the factor.

?Your turn

What is the derivative of f(x) = (x² + 1)³?

A.3·(x² + 1)²B.3·(2x)² = 12x²C.6x·(x² + 1)²D.3·(x² + 1)² · x

Check

?Your turn

Derivative of f(x) = cos(3x + 1)?

A.−sin(3x + 1)B.−3·sin(3x + 1)C.3·sin(3x + 1)D.sin(3x + 1)

Check