The Chain Rule

Main idea

When one function sits inside another, rates of change multiply. Differentiate the outer function, leaving the inside alone, then multiply by the derivative of the inside.

\frac{d}{dx}f\big(g(x)\big) = {\htmlData{sym=0}{f'\big(g(x)\big)}}\cdot {\htmlData{sym=1}{g'(x)}}

What the symbols mean

$f'\big(g(x)\big)$
Outer derivative
Differentiate the outside function, leaving the inside alone.
$g'(x)$
Inner derivative
How fast the inside changes.

Leibniz form

\frac{dy}{dx} = \frac{dy}{d{\htmlData{sym=0}{u}}}\cdot\frac{d{\htmlData{sym=0}{u}}}{dx}

What the symbols mean

$u$
Inside variable
u = g(x). Rates multiply along the chain.

Explanation

The chain rule is simply how you differentiate a function that's inside another function, like $\sin(x^2)$ or $(3x + 1)^5$ . The name comes from thinking of it as a chain of links: $x$ feeds into the inside function, and the inside feeds into the outside. All it really says is that rates multiply along the chain. The formula $f'\big(g(x)\big)\cdot g'(x)$ looks crowded, but it's two pieces: $f'\big(g(x)\big)$ (how fast the outside changes, with the inside left untouched) times $g'(x)$ (how fast the inside changes). That's the whole rule, and it's the one you'll use more than any other.

Rates multiply. Here's why it should be a product. Suppose $y$ changes 3 times as fast as $u$ , and $u$ changes 2 times as fast as $x$ . Nudge $x$ by a little, and $u$ moves twice as much. That makes $y$ move three times as much as $u$ did, so $y$ moves $3 \times 2 = 6$ times as much as $x$ . In Leibniz notation it almost looks like fractions canceling: $\frac{dy}{du}\cdot\frac{du}{dx} = \frac{dy}{dx}$ . (They aren't really fractions, but it's a helpful way to remember it.)

So far, every rule we've had needed the function to be a plain power, exponential or trig function of $x$ . The chain rule is what lets all of them work when $x$ is replaced by something more complicated.

How to apply it. Name the outside and the inside. Differentiate the outside as if the inside were a single variable, then multiply by the inside's derivative.

A common mistake. It's tempting to differentiate the outside and stop, because that's what the old rules trained you to do. So $\frac{d}{dx}\sin(5x)$ comes out as $\cos(5x)$ . It isn't: the inside $5x$ changes 5 times as fast as $x$ , so the answer is $5\cos(5x)$ . Forgetting the inner derivative is the single most common calculus mistake there is.

Patterns worth memorizing:

Function	Derivative
$[g(x)]^n$	$n[g(x)]^{n-1}\,g'(x)$
$e^{g(x)}$	$e^{g(x)}\,g'(x)$
$\ln g(x)$	$\dfrac{g'(x)}{g(x)}$
$\sin g(x)$	$\cos g(x)\cdot g'(x)$

For a chain of three or more, keep peeling layers from the outside in, one rate per layer: $\frac{d}{dx}\sin^3(4x) = 3\sin^2(4x)\cdot\cos(4x)\cdot 4$ .

See it

This is y = sin(x²): sine wrapped around the inner function u = x². Its slope is the outer rate cos(x²) times the inner rate 2x. Notice the waves crowd together as x grows. That's the inner rate 2x getting bigger, so by x = 2.5 the tangent can be as steep as 5 even though sine alone never gets steeper than 1.

Worked examples

Example 1

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

1
The outside is $u^5$ , with derivative $5u^4$ . The inside is $u = x^3 + 2x$ , with $u' = 3x^2 + 2$ . Multiply them, putting the inside back in for $u$ .

$5(x^3 + 2x)^4(3x^2 + 2)$

Example 2

Differentiate $e^{\sin x}$ .

1
The outside is $e^u$ , which is its own derivative, so it stays $e^{\sin x}$ . The inside is $\sin x$ , with derivative $\cos x$ . Multiply the two.

$\cos x\, e^{\sin x}$

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Differentiate $(2x + 1)^4$ .

Problem 2+5 XP

Differentiate $\sqrt{x^2 + 1}$ .

Problem 3+5 XP

Differentiate $\sin(5x)$ .

Problem 4+5 XP

Differentiate $e^{x^2}$ .

Problem 5+5 XP

Differentiate $\ln(x^2 + 4)$ .

Problem 6+5 XP

Differentiate $\cos^3 x$ .

Problem 7+5 XP

A balloon's radius grows as $r(t) = 2 + 0.5t$ cm, and its volume is $V = \frac43\pi r^3$ . How fast is the volume growing at $t = 2$ s?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Differentiate $f(x) = \sqrt{1 + \sin^2(3x)}$ .

Where it's used

Backpropagation, the algorithm that trains neural networks, is the chain rule applied through millions of nested functions. Related rates (two lessons on) is the chain rule with time as the variable.

Gradient DescentCalc III · Computer Science

Derivatives of Hyperbolic Functions Implicit Differentiation

The Chain Rule

Main idea

When one function sits inside another, rates of change multiply. Differentiate the outer function, leaving the inside alone, then multiply by the derivative of the inside.

\frac{d}{dx}f\big(g(x)\big) = {\htmlData{sym=0}{f'\big(g(x)\big)}}\cdot {\htmlData{sym=1}{g'(x)}}

What the symbols mean

$f'\big(g(x)\big)$
Outer derivative
Differentiate the outside function, leaving the inside alone.
$g'(x)$
Inner derivative
How fast the inside changes.

Leibniz form

\frac{dy}{dx} = \frac{dy}{d{\htmlData{sym=0}{u}}}\cdot\frac{d{\htmlData{sym=0}{u}}}{dx}

What the symbols mean

$u$
Inside variable
u = g(x). Rates multiply along the chain.

Explanation

How to apply it. Name the outside and the inside. Differentiate the outside as if the inside were a single variable, then multiply by the inside's derivative.

Patterns worth memorizing:

Function	Derivative
$[g(x)]^n$	$n[g(x)]^{n-1}\,g'(x)$
$e^{g(x)}$	$e^{g(x)}\,g'(x)$
$\ln g(x)$	$\dfrac{g'(x)}{g(x)}$
$\sin g(x)$	$\cos g(x)\cdot g'(x)$

For a chain of three or more, keep peeling layers from the outside in, one rate per layer: $\frac{d}{dx}\sin^3(4x) = 3\sin^2(4x)\cdot\cos(4x)\cdot 4$ .

See it

Worked examples

Example 1

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

1
The outside is $u^5$ , with derivative $5u^4$ . The inside is $u = x^3 + 2x$ , with $u' = 3x^2 + 2$ . Multiply them, putting the inside back in for $u$ .

$5(x^3 + 2x)^4(3x^2 + 2)$

Example 2

Differentiate $e^{\sin x}$ .

1
The outside is $e^u$ , which is its own derivative, so it stays $e^{\sin x}$ . The inside is $\sin x$ , with derivative $\cos x$ . Multiply the two.

$\cos x\, e^{\sin x}$

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Differentiate $(2x + 1)^4$ .

Problem 2+5 XP

Differentiate $\sqrt{x^2 + 1}$ .

Problem 3+5 XP

Differentiate $\sin(5x)$ .

Problem 4+5 XP

Differentiate $e^{x^2}$ .

Problem 5+5 XP

Differentiate $\ln(x^2 + 4)$ .

Problem 6+5 XP

Differentiate $\cos^3 x$ .

Problem 7+5 XP

A balloon's radius grows as $r(t) = 2 + 0.5t$ cm, and its volume is $V = \frac43\pi r^3$ . How fast is the volume growing at $t = 2$ s?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Differentiate $f(x) = \sqrt{1 + \sin^2(3x)}$ .

Where it's used

Backpropagation, the algorithm that trains neural networks, is the chain rule applied through millions of nested functions. Related rates (two lessons on) is the chain rule with time as the variable.

Gradient DescentCalc III · Computer Science