Almost every function that appears outside a textbook is built by composition — $\sqrt{1+x^2}$ , $e^{-t^2/2}$ , $\sin(2\pi f t)$ — and none of the rules so far can differentiate a single one of them. The chain rule is the one tool that handles composition, and once you add it to the sum, product and quotient rules you can differentiate every elementary function you are able to write down. It is also the rule that is misapplied most often, and almost always in exactly one way: by forgetting the derivative of the inside.

Statement of the rule

Theorem (Chain Rule). Suppose $g$ is differentiable at $a$ , and $f$ is differentiable at the point $g(a)$ . Then the composite $F=f\circ g$ , defined by $F(x)=f(g(x))$ , is differentiable at $a$ and

F'(a)=f'\big(g(a)\big)\cdot g'(a).

Read the hypotheses carefully. The outer function $f$ must be differentiable at $g(a)$ — the output of the inner function — not at $a$ . It is not enough for $f$ to be differentiable at $a$ , and it is not enough for the composite to be defined. The last row of the "Common mistakes" table below, and the "Restriction" column of the templates table, show what goes wrong when this hypothesis fails: for instance $f(u)=\sqrt u$ with $g(x)=x^2$ at $a=0$ , where $g$ is differentiable, $g(0)=0$ , and $\sqrt u$ is not differentiable at $u=0$ — and indeed $\sqrt{x^2}=|x|$ has no derivative at $0$ .

Leibniz notation. Introduce an intermediate variable: let $u=g(x)$ and $y=f(u)$ . If $u$ is differentiable at $x$ and $y$ is differentiable at $u$ , then

\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx},

where $\dfrac{dy}{du}$ is evaluated at $u=g(x)$ . That last clause carries all the content of " $f'(g(a))$ " and is exactly what Leibniz notation hides.

Working form. For any differentiable $u=u(x)$ ,

\frac{d}{dx}\big[f(u)\big]=f'(u)\cdot\frac{du}{dx}

— derivative of the outside, inside left untouched, times the derivative of the inside.

Why "cancelling $du$ " is a mnemonic, not a proof

The tempting argument $\dfrac{\Delta y}{\Delta x}=\dfrac{\Delta y}{\Delta u}\cdot\dfrac{\Delta u}{\Delta x}$ needs $\Delta u=g(x)-g(a)\ne 0$ , and $\Delta u$ can vanish at points arbitrarily close to $a$ (take $g(x)=x^2\sin(1/x)$ , $g(0)=0$ ). The repair: because $f$ is differentiable at $b=g(a)$ , define

\varphi(u)=\begin{cases}\dfrac{f(u)-f(b)}{u-b}, & u\ne b\\ f'(b), & u=b\end{cases}

Then $\varphi$ is continuous at $b$ and $f(u)-f(b)=\varphi(u)(u-b)$ for every $u$ , including $u=b$ . So for $x\ne a$ ,

\frac{f(g(x))-f(g(a))}{x-a}=\varphi\big(g(x)\big)\cdot\frac{g(x)-g(a)}{x-a},

an identity that survives $g(x)=g(a)$ . Since $g$ is differentiable at $a$ it is continuous there, so $g(x)\to b$ and $\varphi(g(x))\to\varphi(b)=f'(g(a))$ , while the second factor tends to $g'(a)$ . That is the theorem.

Identifying the inner and the outer function

Ask: if I evaluated this at $x=2$ on a calculator, which key would I press last? That last operation is the outer function; everything pressed before it is the inner function.

Expression	Inner $u$	Outer	Derivative
$(3x^2+1)^5$	$3x^2+1$	$u^5$	$5(3x^2+1)^4\cdot 6x$
$\sin(x^3)$	$x^3$	$\sin u$	$3x^2\cos(x^3)$
$\sin^3 x=(\sin x)^3$	$\sin x$	$u^3$	$3\sin^2x\cos x$
$e^{-5x}$	$-5x$	$e^u$	$-5e^{-5x}$
$\sqrt{4-x^2}$	$4-x^2$	$\sqrt u$	$\dfrac{-x}{\sqrt{4-x^2}}$
$\ln(x^2+1)$	$x^2+1$	$\ln u$	$\dfrac{2x}{x^2+1}$

Rows 2 and 3 are the point of the exercise: $\sin(x^3)$ and $\sin^3x$ are different functions and the roles of inner and outer are swapped.

Standard templates

Throughout, $u=u(x)$ is differentiable and $u'=\dfrac{du}{dx}$ .

Function	Derivative	Restriction
$u^n$	$nu^{n-1}u'$	$n$ a negative integer: need $u\ne 0$ ; $n$ not an integer: need $u>0$
$\sqrt{u}$	$\dfrac{u'}{2\sqrt u}$	$u>0$ (not differentiable where $u=0$ )
$\dfrac{1}{u}$	$-\dfrac{u'}{u^{2}}$	$u\ne 0$
$e^{u}$	$e^{u}u'$	—
$a^{u}$	$a^{u}(\ln a)\,u'$	$a>0$
$\ln u$	$\dfrac{u'}{u}$	$u>0$
$\log_a u$	$\dfrac{u'}{u\ln a}$	$u>0$ , $a>0$ , $a\ne 1$
$\sin u$	$(\cos u)\,u'$	—
$\cos u$	$-(\sin u)\,u'$	—
$\tan u$	$(\sec^2 u)\,u'$	$u\ne\frac{\pi}{2}+k\pi$
$\sec u$	$(\sec u\tan u)\,u'$	$u\ne\frac{\pi}{2}+k\pi$
$\cot u$	$-(\csc^2 u)\,u'$	$u\ne k\pi$
$\csc u$	$-(\csc u\cot u)\,u'$	$u\ne k\pi$
$f(ax+b)$	$a\,f'(ax+b)$	the workhorse special case

Worked example 1 — plain composition

Differentiate $f(x)=(3x^2+1)^5$ and find $f'(1)$ .

Outer: fifth power. Inner: $u=3x^2+1$ , so $u'=6x$ .

\begin{aligned} f'(x) &= 5u^{4}\cdot u' \\ &= 5(3x^2+1)^4\cdot 6x \\ &= 30x(3x^2+1)^4 . \end{aligned}

At $x=1$ : $3(1)^2+1=4$ and $4^4=256$ , so $f'(1)=30(1)(256)=7680$ .

Check. Expanding the fifth power is impractical, so use a centred difference with $h=0.001$ : $f(1.001)\approx 1031.707$ , $f(0.999)\approx 1016.347$ , and $\dfrac{1031.707-1016.347}{0.002}\approx 7680$ . Agreed.

Worked example 2 — product rule and chain rule together

Differentiate $y=x^2\sqrt{1-4x}$ . (Defined for $x\le\frac14$ ; differentiable for $x<\frac14$ , since $\sqrt{u}$ fails to be differentiable at $u=0$ .)

The outermost operation is a product, so the product rule fires first; the chain rule is then needed inside it.

\begin{aligned} y' &= (x^2)'\sqrt{1-4x}+x^2\left(\sqrt{1-4x}\right)' \\ &= 2x\sqrt{1-4x}+x^2\cdot\frac{-4}{2\sqrt{1-4x}} \\ &= 2x\sqrt{1-4x}-\frac{2x^2}{\sqrt{1-4x}} \\ &= \frac{2x(1-4x)-2x^2}{\sqrt{1-4x}} \\ &= \frac{2x-8x^2-2x^2}{\sqrt{1-4x}}=\frac{2x(1-5x)}{\sqrt{1-4x}} . \end{aligned}

Check at $x=-2$ . Final formula: $\dfrac{2(-2)\big(1-5(-2)\big)}{\sqrt{9}}=\dfrac{(-4)(11)}{3}=-\dfrac{44}{3}$ . Third line, computed independently: $2(-2)(3)-\dfrac{2(4)}{3}=-12-\dfrac{8}{3}=-\dfrac{44}{3}$ . The two match.

Repeated (nested) application

For three layers, $(f\circ g\circ h)'(x)=f'\big(g(h(x))\big)\cdot g'\big(h(x)\big)\cdot h'(x)$ , or in Leibniz form $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dv}\cdot\dfrac{dv}{dx}$ . Peel from the outside in; every layer contributes exactly one factor, and you stop when what remains inside is $x$ itself.

Take $y=\sin^3(4x)$ — three layers: cube, sine, $4x$ .

\begin{aligned} y' &= 3\big[\sin(4x)\big]^{2}\cdot\frac{d}{dx}\sin(4x) \\ &= 3\sin^2(4x)\cdot\cos(4x)\cdot\frac{d}{dx}(4x) \\ &= 3\sin^2(4x)\cos(4x)\cdot 4 \\ &= 12\sin^2(4x)\cos(4x). \end{aligned}

At $x=\frac{\pi}{12}$ we have $4x=\frac{\pi}{3}$ , $\sin\frac{\pi}{3}=\frac{\sqrt3}{2}$ , $\cos\frac{\pi}{3}=\frac12$ , so $y'=12\cdot\frac34\cdot\frac12=\frac92$ . A centred difference at $x=\frac{\pi}{12}$ with $h=0.001$ gives $4.49992$ , confirming $\frac92$ .

Which rule fires first

Look at the outermost operation of the whole expression:

Outermost operation	Rule that fires first
sum or difference	sum rule, then treat each term separately
product of two non-constant factors	product rule
a fraction	quotient rule, or rewrite as a product with a negative power
a function applied to something	chain rule

So $\left(\dfrac{3x-1}{x+2}\right)^4$ is a fourth power on the outside: chain rule first, quotient rule inside. But $\dfrac{(3x-1)^4}{x+2}$ is a fraction on the outside: quotient rule first, chain rule inside.

Common mistakes

Wrong	Right	What went missing
$\frac{d}{dx}(5x+1)^4=4(5x+1)^3$	$20(5x+1)^3$	the inner derivative $u'=5$
$\frac{d}{dx}\sin(3x)=\cos(3x)$	$3\cos(3x)$	the inner derivative $u'=3$
$\frac{d}{dx}e^{x^2}=e^{x^2}$	$2x\,e^{x^2}$	the inner derivative $u'=2x$
$\frac{d}{dx}\sqrt{x^2+1}=\dfrac{1}{2\sqrt{x^2+1}}$	$\dfrac{x}{\sqrt{x^2+1}}$	the inner derivative $u'=2x$
$\frac{d}{dx}\ln(3x)=\dfrac{1}{3x}$	$\dfrac{3}{3x}=\dfrac{1}{x}$	the inner derivative $u'=3$
$\frac{d}{dx}\sin(x^2)=2x\cos x$	$2x\cos(x^2)$	the outer derivative must be evaluated at $u=x^2$ , not at $x$
$\frac{d}{dx}f(g(x))=f'(g'(x))$	$f'(g(x))\,g'(x)$	$f'$ is evaluated at $g(x)$ and multiplied by $g'(x)$
$\frac{d}{dx}(x^2+1)^3=3(x^2+1)^2+2x$	$3(x^2+1)^2(2x)$	the chain rule multiplies; it never adds
$\frac{d}{dx}e^{5x}=5x\,e^{5x-1}$	$5e^{5x}$	the power rule does not apply to a variable exponent
$\frac{d}{dx}\cos^2x=-\sin^2x$	$-2\cos x\sin x=-\sin(2x)$	$\cos^2x$ means $(\cos x)^2$ ; the outer function is the square
$\frac{d}{dx}\big[x\sin(2x)\big]=\sin(2x)+x\cos(2x)$	$\sin(2x)+2x\cos(2x)$	chain rule inside the product rule
stopping at $\frac{dy}{du}$	multiply by $\frac{du}{dx}$ , then substitute $u=g(x)$	the answer must be a function of $x$
$\frac{d}{dx}(x^2-1)^{1/2}=\dfrac{x}{\sqrt{x^2-1}}$ for every $x$	only for $x>1$ or $x<-1$	a non-integer power needs a positive base

One structural error deserves its own line, because it is the single most common: the missing inner derivative. Whenever your answer to $\frac{d}{dx}f(u)$ contains no factor of $u'$ at all, and $u$ is not simply $x$ , you have made it.

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$1$	$3$	$4$	$2$	$5$
$2$	$1$	$6$	$3$	$-2$
$3$	$2$	$7$	$1$	$8$

Navigation

The Chain Rule

Statement of the rule

Why "cancelling $du$ " is a mnemonic, not a proof

Identifying the inner and the outer function

Standard templates

Worked example 1 — plain composition

Worked example 2 — product rule and chain rule together

Repeated (nested) application

Which rule fires first

Common mistakes

Key terms

Practice Problems

Navigation

Statement of the rule

Why "cancelling dududu" is a mnemonic, not a proof

Identifying the inner and the outer function

Standard templates

Worked example 1 — plain composition

Worked example 2 — product rule and chain rule together

Repeated (nested) application

Which rule fires first

Common mistakes

Key terms

Practice Problems

Why "cancelling $du$ " is a mnemonic, not a proof