Math · Calculus I · Concept

The chain rule for derivatives

The chain rule differentiates a function inside another function, such as $sin (x^{2})$ or ${(3 x + 1)}^{50}$ . Differentiate the outside function while leaving the inside untouched, then multiply by the derivative of the inside. In symbols, the derivative of $f (g (x))$ is $f^{'} (g (x)) \cdot g^{'} (x)$ .

Before this

The limit definition of the derivative. A derivative is a rate. It tells you how many times as fast the output changes as the input, at one point.
Derivatives of $e^{x}$ , $ln x$ and trig functions. The derivatives of $e^{x}$ , $ln x$ , $sin x$ and $cos x$ are $e^{x}$ , $1 / x$ , $cos x$ and $- sin x$ , and with the power rule they supply the rates that the chain rule multiplies together.
Composite and inverse functions. $f (g (x))$ means you work out $g (x)$ first and then apply $f$ to the result.

A concrete case

Follow one small change

Take $y = {(3 x + 1)}^{2}$ . Working out $y$ for a given $x$ happens in two stages. First you find $3 x + 1$ , then you square the result. Call the first stage the inside and the second the outside.

Start at $x = 1$ . The inside is $3 (1) + 1 = 4$ , and $y$ is $4^{2} = 16$ . Now nudge $x$ up by $0.01$ , to $1.01$ . The inside becomes $4.03$ . It moved $0.03$ , three times as far as $x$ moved. Then $y$ becomes ${4.03}^{2} = 16.2409$ . It moved $0.2409$ , about eight times as far as the inside moved.

Put the two stages together. $y$ moved about $3 \times 8 = 24$ times as far as $x$ did, and $0.2409 \div 0.01 = 24.09$ confirms it. The exact rate of change of $y$ at $x = 1$ is $24$ . Each stage multiplied the change by its own rate, and the two rates multiplied. That is the whole idea of the chain rule.

\frac{Δ y}{Δ x} = \frac{0.2409}{0.01} = 24.09

Following a nudge of $0.01$ through $y$ $= {(3 x + 1)}^{2}$ at $x = 1$
Quantity	At $x = 1$	At $x = 1.01$	Change	Compared with the stage before
$x$	$1$	$1.01$	$0.01$	the nudge
$u$ $= 3 x + 1$	$4$	$4.03$	$0.03$	$3$ times the change in $x$
$y = u^{2}$	$16$	$16.2409$	$0.2409$	about $8$ times the change in $u$

The rule

Functions built in two stages

The power, product and quotient rules handle functions that are added, multiplied or divided. The chain rule handles the remaining way to build a function, which is to apply one function to the output of another. This is called composition, and the result is a composite function, written $f (g (x))$ .

$sin (x^{2})$ is not $sin x$ times $x^{2}$ . The $x^{2}$ sits inside the sine, and none of the other rules can reach it. ${(3 x + 1)}^{50}$ could be expanded into $51$ terms, but the chain rule differentiates it in one line.

To tell the inside from the outside, work out the value for one particular $x$ , say $x = 2$ . Whatever you work out first is the inside. The last operation you apply is the outside. For $sin (x^{2})$ , you square $2$ to get $4$ and then take the sine, so the inside is $x^{2}$ and the outside is the sine. For ${(3 x + 1)}^{5}$ , you work out $3 \cdot 2 + 1 = 7$ and then raise $7$ to the 5th power, so the inside is $3 x + 1$ and the outside is the 5th power.

Give the inside its own letter, usually $u$ . Writing $u = g (x)$ and $y = f (u)$ splits one complicated function into two simple ones, and each simple one is easy to differentiate.

sin (x^{2}) {(3 x + 1)}^{50} e^{- x^{2}} \sqrt{x^{2} + 1}

\begin{array}{c} sin (x^{2}) {(3 x + 1)}^{50} \\ e^{- x^{2}} \sqrt{x^{2} + 1} \end{array}

The inside and the outside of common composite functions
Function	Inside, $u$	Outside, in terms of $u$
${(3 x + 1)}^{5}$	$3 x + 1$	$u^{5}$
$sin (x^{2})$	$x^{2}$	$sin u$
${sin}^{2} x$ , which means ${(sin x)}^{2}$	$sin x$	$u^{2}$
$e^{4 x}$	$4 x$	$e^{u}$
$\sqrt{x^{2} + 1}$	$x^{2} + 1$	$\sqrt{u}$
$ln (cos x)$	$cos x$	$ln u$
$1 / {(2 x - 5)}^{3}$	$2 x - 5$	$1 / u^{3}$ , which is $u^{- 3}$

The chain rule

If $y = f (u)$ and $u = g (x)$ , the rate of $y$ with respect to $x$ is the rate of $y$ with respect to $u$ times the rate of $u$ with respect to $x$ . In Leibniz notation this is $d y / d x = (d y / d u) (d u / d x)$ .

In function notation the rule says three things. Take the derivative of the outside. Evaluate it at the inside. Multiply by the derivative of the inside. “Evaluate it at the inside” means you find the outside’s derivative and then put the whole inside expression where its variable was. $f^{'} (g (x))$ means $f^{'}$ applied to $g (x)$ . It does not mean $f^{'} (x)$ times $g (x)$ .

\frac{d}{d x} f (g (x)) = f^{'} (g (x)) g^{'} (x)

\begin{aligned} \frac{d}{d x} f (g (x)) \\ = f^{'} (g (x)) g^{'} (x) \end{aligned}

Why it works

Why the rates multiply

A derivative is a rate. It says how many times as fast the output changes as the input. In the nudge above, the inside changed $3$ times as fast as $x$ , because $d u / d x = 3$ . The outside changed about $8$ times as fast as the inside, because $d y / d u = 2 u$ , which is $8$ when $u = 4$ . A change in $x$ passes through both stages, and each stage scales it by its own rate. The result is $3 \times 8 = 24$ times the change in $x$ .

Gears make the same point. If the middle gear turns $3$ times as fast as the first, and the last turns $8$ times as fast as the middle, then the last turns $24$ times as fast as the first. Nobody adds gear ratios. They multiply.

The nudge gave $24.09$ rather than $24$ because $0.01$ is a small change but not an infinitely small one. A nudge of $0.001$ gives $24.009$ , and as the nudge shrinks the ratio closes in on $24$ exactly.

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} = 8 \cdot 3 = 24

The method

The method in five steps

Until the chain rule feels automatic, write every stage down. The table works through $y = sin (x^{2})$ .

The finished derivative contains only $x$ . If a $u$ is still there, the last step isn’t finished.

\frac{d y}{d x} = cos (x^{2}) \cdot 2 x = 2 x cos (x^{2})

The five steps of the chain rule on $y$ $= sin (x^{2})$
Step	What to do	For $y$ $= sin (x^{2})$
1	Name the inside	$u = x^{2}$
2	Rewrite $y$ using $u$	$y$ $= sin u$
3	Differentiate the outside with respect to $u$	$d y / d u$ $= cos u$
4	Differentiate the inside with respect to $x$	$d u / d x$ $= 2 x$
5	Multiply, then put the inside back in place of $u$	$d y / d x$ $= cos (x^{2}) \cdot 2 x$ $= 2 x cos (x^{2})$

The shortcut without $u$

With practice you can skip the letter $u$ . Differentiate the outside as if the inside were a single letter, copying the inside exactly as it is. Then multiply by the derivative of the inside.

For ${(x^{2} + 5)}^{7}$ , the outside is the 7th power, so write $7 {(x^{2} + 5)}^{6}$ with the inside copied unchanged. The derivative of the inside is $2 x$ . Multiply to get $14 x {(x^{2} + 5)}^{6}$ .

The table lists the patterns you will use most. Each one is the chain rule with a familiar outside function.

\frac{d}{d x} {(x^{2} + 5)}^{7} = 7 {(x^{2} + 5)}^{6} \cdot 2 x = 14 x {(x^{2} + 5)}^{6}

\begin{aligned} \frac{d}{d x} {(x^{2} + 5)}^{7} \\ = 7 {(x^{2} + 5)}^{6} \cdot 2 x \\ = 14 x {(x^{2} + 5)}^{6} \end{aligned}

Chain rule patterns
Function	Derivative
${[g (x)]}^{n}$	$n {[g (x)]}^{n - 1} \cdot g^{'} (x)$
$\sqrt{g (x)}$	$g^{'} (x) / (2 \sqrt{g (x)})$
$e^{g (x)}$	$e^{g (x)} \cdot g^{'} (x)$
$ln g (x)$	$g^{'} (x) / g (x)$
$sin g (x)$	$cos g (x) \cdot g^{'} (x)$
$cos g (x)$	$- sin g (x) \cdot g^{'} (x)$
$tan g (x)$	${sec}^{2} g (x) \cdot g^{'} (x)$

An inside of $k x$ , or just $x$

When the inside is a number times $x$ , its derivative is that number, and the chain rule multiplies by it. The derivative of $sin (5 x)$ is $5 cos (5 x)$ , and the derivative of $e^{3 x}$ is $3 e^{3 x}$ .

The chain rule is at work even when the inside is plain $x$ . The derivative of $sin x$ is $cos x$ times the derivative of $x$ , which is $1$ , and a factor of $1$ is invisible. That is why the basic rules work without it, and why an extra factor appears as soon as the inside is anything other than $x$ .

\frac{d}{d x} sin (5 x) = 5 cos (5 x) \frac{d}{d x} e^{3 x} = 3 e^{3 x}

\begin{array}{c} \frac{d}{d x} sin (5 x) = 5 cos (5 x) \\ \frac{d}{d x} e^{3 x} = 3 e^{3 x} \end{array}

Layers inside layers

Some functions have three or more layers. Peel them from the outside in, and write one derivative factor for each layer, copying everything inside that layer unchanged.

In $y = {sin}^{3} (4 x)$ , which means ${(sin 4 x)}^{3}$ , the outermost layer is the cube, the middle layer is the sine and the innermost layer is $4 x$ . The cube gives $3 {sin}^{2} (4 x)$ , the sine gives $cos (4 x)$ , and $4 x$ gives $4$ . Multiply all three to get $12 {sin}^{2} (4 x) cos (4 x)$ .

Count the layers before you start. The number of factors you multiply should match.

\frac{d}{d x} {sin}^{3} (4 x) = 3 {sin}^{2} (4 x) \cdot cos (4 x) \cdot 4 = 12 {sin}^{2} (4 x) cos (4 x)

\begin{aligned} \frac{d}{d x} {sin}^{3} (4 x) \\ = 3 {sin}^{2} (4 x) \cdot cos (4 x) \cdot 4 \\ = 12 {sin}^{2} (4 x) cos (4 x) \end{aligned}

Using it with the product and quotient rules

When a problem needs several rules, find the structure at the top level first by asking which operation is done last. In $x^{2} e^{3 x}$ , you work out $x^{2}$ and $e^{3 x}$ separately and then multiply. The function is a product, and the product rule comes first. The chain rule appears inside it, when you differentiate the factor $e^{3 x}$ .

If the last operation is a function applied to an expression, start with the chain rule instead. In $sin (x e^{x})$ , the sine is applied last. The outside is the sine, and its inside, $x e^{x}$ , needs the product rule.

\frac{d}{d x} (x^{2} e^{3 x}) = 2 x e^{3 x} + x^{2} \cdot 3 e^{3 x} = x e^{3 x} (2 + 3 x)

\begin{aligned} \frac{d}{d x} (x^{2} e^{3 x}) \\ = 2 x e^{3 x} + x^{2} \cdot 3 e^{3 x} \\ = x e^{3 x} (2 + 3 x) \end{aligned}

Where it breaks

The chain rule needs both stages to have a derivative where they are used. The inside must be differentiable at $x$ , and the outside must be differentiable at the value the inside produces. If either fails, the rule gives no answer there.

Take $f (x) = \sqrt{x^{2}}$ . The rule gives $f^{'} (x)$ $= 2 x / (2 \sqrt{x^{2}})$ $= x / \sqrt{x^{2}}$ , which is $1$ for $x > 0$ and $- 1$ for $x < 0$ . At $x = 0$ the inside is $0$ , and the square root has no derivative at $0$ , because its graph is vertical there. The formula breaks down at $x = 0$ , and so does the derivative. $\sqrt{x^{2}}$ is $∣ x ∣$ , whose graph has a corner at the origin.

The patterns for $\sqrt{g (x)}$ and $ln g (x)$ need $g (x) > 0$ for the same reason. The derivative of $ln (x^{2} - 4)$ is $2 x / (x^{2} - 4)$ , but only where $x^{2} - 4 > 0$ , which means $x < - 2$ or $x > 2$ .

The rule also has a shape it fits. It is for a function applied to a function. A product such as $x^{2} sin x$ is not a composition. You work out $x^{2}$ and $sin x$ separately and multiply, so it needs the product rule. ${sin}^{2} x$ and $sin (x^{2})$ are both compositions, built from the same two pieces in opposite orders, and their derivatives differ.

\frac{d}{d x} \sqrt{x^{2}} = \frac{x}{\sqrt{x^{2}}} = {\begin{cases} 1, & x > 0 \\ - 1, & x < 0 \end{cases}

\begin{aligned} \frac{d}{d x} \sqrt{x^{2}} = \frac{x}{\sqrt{x^{2}}} \\ = {\begin{cases} 1, & x > 0 \\ - 1, & x < 0 \end{cases} \end{aligned}

Check your answer

A chain rule derivative most often goes wrong by losing a layer, so test for that first.

Count the factors. A function with two layers needs two factors, and one with three layers needs three.

Try a number. Pick a small $h$ such as $0.001$ . The ratio $[y (x + h) - y (x)] / h$ should be close to your derivative at that $x$ .

Differentiate another way when you can. ${(3 x + 1)}^{2}$ expands to $9 x^{2} + 6 x + 1$ , and the power rule gives $18 x + 6$ , the same as the chain rule.

To have Chalk check the work, write it in a Math box. Put the function on the first row as $f (x) = \dots$ and each step of your derivative on its own row below it, $f^{'} (x) = \dots$ . The Checker marks each row ✓ or ✗ against the row above. Name the function with the variable its formula uses, so a formula in $r$ goes in as $f (r)$ and $f^{'} (r)$ . For $y = \dots$ write the derivative as $d y / d x = \dots$ .

A “?” means Chalk could not follow that row. It is not a ✗, and the work may still be right. Explore on a Math box row, then Differentiate, shows the outside-in steps for a function you type.

Why the rule is true, and where it leads

Leibniz notation makes the chain rule look like cancelling $d u$ . That is a good memory aid but not a proof, since $d y / d u$ and $d u / d x$ are limits, not fractions.

For an actual small change $Δ x$ , the changes do satisfy $Δ y / Δ x = (Δ y / Δ u) (Δ u / Δ x)$ whenever $Δ u \neq 0$ . Letting $Δ x$ shrink to $0$ turns the three ratios into the three derivatives. A complete proof must also handle an inside that stops changing near the point, where $Δ u = 0$ and the middle ratio is undefined. It does this by writing each function as its linear approximation plus an error that shrinks faster than the change.

The chain rule is behind much of what comes next. Implicit differentiation uses it to differentiate $y^{2}$ as $2 y \cdot d y / d x$ . Related rates uses it to link quantities that change together in time. Read backwards, it becomes u-substitution, the first method for undoing a derivative.

\frac{Δ y}{Δ x} = \frac{Δ y}{Δ u} \cdot \frac{Δ u}{Δ x} (Δ u \neq 0)

Common mistakes

Forgetting the derivative of the inside is the most common slip. The derivative of ${(3 x + 1)}^{2}$ is $2 (3 x + 1) \cdot 3 = 18 x + 6$ , not $2 (3 x + 1)$ , because the inside changes $3$ times as fast as $x$ .
Changing the inside while differentiating the outside. The derivative of $sin (x^{2})$ is $cos (x^{2}) \cdot 2 x$ . It is not $cos (2 x)$ and not $cos x \cdot 2 x$ , because the outside’s derivative is taken at the inside, which stays exactly as it was.
Reading ${sin}^{2} x$ as $sin (x^{2})$ . ${sin}^{2} x$ means ${(sin x)}^{2}$ , whose outside is the square, so its derivative is $2 sin x cos x$ . The derivative of $sin (x^{2})$ is $2 x cos (x^{2})$ .
Stopping after one layer. ${sin}^{3} (4 x)$ has three layers, so its derivative needs three factors, $3 {sin}^{2} (4 x) \cdot cos (4 x) \cdot 4$ . Leaving out the $4$ drops the innermost layer’s rate.
Using the chain rule on a product. $x^{2} sin x$ is $x^{2}$ times $sin x$ , not one function inside another. It needs the product rule, which gives $2 x sin x + x^{2} cos x$ .
Leaving $u$ in the answer. After multiplying, replace $u$ with the inside, so that the derivative is written in terms of $x$ alone.

Key terms

Chain rule: The rule for differentiating a function inside another function: ${[f (g (x))]}^{'}$ $= f^{'} (g (x)) \cdot g^{'} (x)$ . Differentiate the outside, keeping the inside as it is, then multiply by the derivative of the inside.
Composition of functions: Feeding one function’s output into another, written $f (g (x))$ or $(f \circ g) (x)$ . The inner output has to be an allowed input for the outer function.
Derivative: The instantaneous rate of change of a function: the limit of the average rate of change as the step shrinks to zero, when that limit exists. On a graph it is the slope of the tangent line.
Product rule: To differentiate a product, take the derivative of the first factor times the second, plus the first times the derivative of the second: ${(u v)}^{'} = u^{'} v + u v^{'}$ .