The Chain Rule: When and How to Apply It (with Examples)

The chain rule is the single most-used tool in differentiation, and also the single biggest source of mistakes. Once you internalise the "outer-then-inner" pattern, you can differentiate almost any composite function in three lines. This guide shows you the pattern, walks through seven escalating examples, and lists the four mistakes worth memorising in advance.

What the chain rule says

If $f$ and $g$ are differentiable, the derivative of the composition $f(g(x))$ is

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

In words: differentiate the outer function evaluated at the inner, then multiply by the derivative of the inner. The "outer" and "inner" labels are non-negotiable — confusing them flips the answer.

A useful mnemonic: the chain rule is "the outer derivative times the inner derivative", never plus, never just one.

Worked examples (easy → hard)

Example 1: $\frac{d}{dx}\sin(2x)$

• Outer: $\sin(u)$, inner: $u = 2x$.
• $\frac{d}{du}\sin(u) = \cos(u)$, $\frac{d}{dx}(2x) = 2$.
• Result: $\cos(2x) \cdot 2 = 2\cos(2x)$.

Example 2: $\frac{d}{dx} e^{x^2}$

• Outer: $e^u$, inner: $u = x^2$.
• $\frac{d}{du} e^u = e^u$, $\frac{d}{dx}(x^2) = 2x$.
• Result: $e^{x^2} \cdot 2x = 2x e^{x^2}$.

Example 3: $\frac{d}{dx}(3x^2 + 1)^4$

• Outer: $u^4$, inner: $u = 3x^2 + 1$.
• $\frac{d}{du} u^4 = 4u^3$, $\frac{d}{dx}(3x^2 + 1) = 6x$.
• Result: $4(3x^2 + 1)^3 \cdot 6x = 24x(3x^2 + 1)^3$.

Example 4: $\frac{d}{dx}\ln(\cos x)$

• Outer: $\ln u$, inner: $u = \cos x$.
• $\frac{d}{du}\ln u = \frac{1}{u}$, $\frac{d}{dx}\cos x = -\sin x$.
• Result: $\frac{1}{\cos x} \cdot (-\sin x) = -\tan x$.

Example 5: $\frac{d}{dx}\sqrt{x^2 + 1}$

• Rewrite as $(x^2 + 1)^{1/2}$.
• Outer: $u^{1/2}$, inner: $u = x^2 + 1$.
• Outer derivative: $\frac{1}{2}u^{-1/2}$. Inner: $2x$.
• Result: $\frac{1}{2}(x^2+1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+1}}$.

Example 6: Nested chain — $\frac{d}{dx}\sin(\cos(x^2))$

Three layers — apply chain rule twice.

• Outermost: $\sin(u)$, inner $u = \cos(x^2)$.
• $\frac{du}{dx} = -\sin(x^2) \cdot 2x$ (chain rule on $\cos(x^2)$).
• Result: $\cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot 2x = -2x\sin(x^2)\cos(\cos(x^2))$.

Example 7: Chain + product rule together — $\frac{d}{dx}\bigl(x^2 \sin(3x)\bigr)$

• Use the product rule first: $(fg)' = f'g + fg'$.
• $f = x^2$, $f' = 2x$. $g = \sin(3x)$, by chain rule $g' = 3\cos(3x)$.
• Result: $2x \sin(3x) + x^2 \cdot 3\cos(3x) = 2x\sin(3x) + 3x^2\cos(3x)$.

The four mistakes worth memorising

1. Forgetting the inner derivative. Writing $\frac{d}{dx}\sin(2x) = \cos(2x)$ is the single most common chain-rule error. The factor of $2$ is mandatory.
2. Differentiating the inner before substituting. $\frac{d}{dx}(3x^2+1)^4$ is not $4(6x)^3$. The outer derivative is evaluated at the inner expression, not at the inner derivative.
3. Mistaking nested function for product. $\sin(2x)$ is a composition, not a product. Use the chain rule, not the product rule.
4. Bracketing trig powers wrong. $\sin^2(x) = (\sin x)^2$ — outer is $u^2$, inner is $\sin x$. Easily confused with $\sin(x^2)$ where outer is $\sin$ and inner is $x^2$.

When you're stuck: the substitution trick

Set $u = \text{(the inner part)}$, find $\frac{dy}{du}$ and $\frac{du}{dx}$, multiply. Even when the function looks intimidating, this rote substitution always works.

Try it yourself

Paste any composite function into our free Derivative Calculator and watch each chain-rule application happen step-by-step. Pair it with our chain rule cheat sheet section for quick reference during homework.

For deeper related material:

• Limit Calculator — limits feed into the chain rule's foundation
• Integral Calculator — substitution is the chain rule run backwards
• Worked example: derivative of sin(2x)
• Worked example: derivative of e^(2x)