The Chain Rule, calculus lesson

If a gear turns three times as fast as a second gear, and the second turns twice as fast as a third, then the first turns six times as fast as the third. Rates multiply along a chain. The chain rule says exactly this for functions built one inside another, and it is the rule you will use more than any other.

Key terms

Composition

Definition The composition of $f$ and $g$ is the function $f \circ g$ given by

(f \circ g)(x) = f(g(x)).

Its domain is the set of all $x$ in the domain of $g$ for which $g(x)$ lies in the domain of $f$ .

In plain words Apply $g$ first, then feed the result into $f$ . An input is allowed only if both steps make sense.

Example With $g(x) = 1 - x^2$ and $f(u) = \sqrt{u}$ , $(f \circ g)(x) = \sqrt{1 - x^2}$ . We need $1 - x^2 \ge 0$ , so the domain is $[-1, 1]$ . In the other order, $(g \circ f)(x) = 1 - (\sqrt{x})^2 = 1 - x$ with domain $[0, \infty)$ , a different function.

Inner and outer functions

Definition When a function is written as $y = f(g(x))$ , the function $g$ is the inner function and $f$ is the outer function. It is common to name the intermediate value, $u = g(x)$ , so that $y = f(u)$ . The same function can usually be split in more than one way, and any split into differentiable pieces works.

In plain words The inner function is what you compute first. The outer function is the last operation you would do with a calculator.

Example For $y = (3x^2 + 1)^5$ the inner function is $u = 3x^2 + 1$ and the outer function is $y = u^5$ . For $y = \sqrt{x^3 + 4}$ the inner function is $u = x^3 + 4$ and the outer is $y = \sqrt{u}$ .

Derivative of a composite

Definition If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then $f \circ g$ is differentiable at $x$ and

(f \circ g)'(x) = f'(g(x)) \cdot g'(x).

In Leibniz notation, with $y = f(u)$ and $u = g(x)$ ,

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

where $\frac{dy}{du}$ is evaluated at $u = g(x)$ .

In plain words Differentiate the outer function, leaving the inner one inside it untouched, then multiply by the derivative of the inner function.

Example For $y = (3x^2 + 1)^5$ ,

\frac{dy}{dx} = 5(3x^2 + 1)^4 \cdot 6x = 30x(3x^2 + 1)^4.

At $x = 1$ this is $30 \cdot 4^4 = 7680$ . Expanding $(3x^2 + 1)^5$ first would give the same answer with much more work.

Nested compositions

Definition For three functions, if $h$ is differentiable at $x$ , $g$ is differentiable at $h(x)$ and $f$ is differentiable at $g(h(x))$ , then

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

Longer chains work the same way, with one factor for each layer.

In plain words Peel the layers from the outside in. Each layer contributes its derivative, evaluated at whatever sits inside it.

Example Let $y = \sqrt{1 + (x^2 + 1)^3}$ . The layers are $\sqrt{\,\cdot\,}$ , then $1 + (\,\cdot\,)^3$ , then $x^2 + 1$ . So

\frac{dy}{dx} = \frac{1}{2\sqrt{1 + (x^2 + 1)^3}} \cdot 3(x^2 + 1)^2 \cdot 2x.

At $x = 1$ we have $x^2 + 1 = 2$ and $1 + 2^3 = 9$ , so the slope there is

\frac{1}{2 \cdot 3} \cdot 3 \cdot 2^2 \cdot 2 = \frac{1}{6} \cdot 12 \cdot 2 = 4.

Rates with intermediate variables

Definition If a quantity $y$ is a differentiable function of $u$ , and $u$ is a differentiable function of time $t$ , then $y$ changes with $t$ at the rate

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

with $\frac{dy}{du}$ evaluated at the current value of $u$ . The units match, since units of $y$ per $u$ times units of $u$ per $t$ give units of $y$ per $t$ .

In plain words If you know how $y$ responds to $u$ and how fast $u$ is changing, multiply to get how fast $y$ is changing.

Example A spherical balloon has volume $V = \frac{4}{3}\pi r^3$ , so $\frac{dV}{dr} = 4\pi r^2$ . If the radius grows at $2$ centimeters per second, then at the moment $r = 3$ centimeters,

\frac{dV}{dt} = 4\pi \cdot 3^2 \cdot 2 = 72\pi \approx 226 \text{ cubic centimeters per second}.

Differentiability assumptions

Definition The chain rule needs $g$ differentiable at $x$ and $f$ differentiable at the point $g(x)$ , not at $x$ . These conditions are sufficient but not necessary. If one of them fails, the composite may or may not be differentiable, and the formula cannot be used to decide. A careful proof also has to handle points where $g(x + h) = g(x)$ for small $h \neq 0$ , which is why the tempting argument of cancelling $\Delta u$ in $\frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}$ is not a complete proof.

In plain words Check that the inner function is smooth at your point and the outer function is smooth where the inner one lands. If not, go back to the limit definition.

Example Take $f(u) = \lvert u \rvert$ and $g(x) = x^2 - 1$ at $x = 1$ . Here $g(1) = 0$ and $f$ is not differentiable at $0$ , so the chain rule does not apply. In fact $\lvert x^2 - 1 \rvert$ has one-sided slopes $-2$ and $2$ at $x = 1$ , a corner. On the other hand, with $f(u) = u^2$ and $g(x) = \lvert x \rvert$ the inner function fails at $0$ , yet $f(g(x)) = x^2$ is differentiable there. The chain rule is silent in both cases, and the answers differ.

Check yourself

Differentiate $y = \sqrt{x^3 + 4}$ and find the slope at $x = 2$ .

Show the answer

With inner function $u = x^3 + 4$ the chain rule gives

y' = \frac{3x^2}{2\sqrt{x^3 + 4}}.

At $x = 2$ the slope is

\frac{12}{2 \sqrt{12}} = \frac{6}{\sqrt{12}} = \sqrt{3}.

Suppose $g(1) = 3$ , $g'(1) = -2$ and $f'(3) = 5$ . Find $(f \circ g)'(1)$ .

Show the answer

The chain rule gives

(f \circ g)'(1) = f'(g(1)) \cdot g'(1) = f'(3) \cdot (-2) = -10.

Differentiate $y = \frac{1}{(2x - 5)^3}$ .

Show the answer

Write $y = (2x - 5)^{-3}$ . For $x \neq \frac{5}{2}$ the chain rule gives

y' = -3(2x - 5)^{-4} \cdot 2 = -\frac{6}{(2x - 5)^4}.

The Chain Rule

Theory, part 49

Practice, part 50

Key terms

Composition

Inner and outer functions

Derivative of a composite

Nested compositions

Rates with intermediate variables

Differentiability assumptions

Check yourself

How this lesson connects

Builds on

Used later in