Math Core

You can now differentiate powers, sums, products, quotients and the basic trig, exponential and log functions. But most functions you meet are built by plugging one function into another, like $\sin(x^2)$ or $\sqrt{3x + 1}$ . The chain rule handles every one of these, which makes it the most-used rule in the whole course.

Functions inside functions

A composite function $h(x) = f(g(x))$ does two jobs in order: first the inner function $g$ acts on $x$ , then the outer function $f$ acts on the result. For example,

function	inner $g(x)$	outer $f(u)$
$(3x^2 + 1)^5$	$3x^2 + 1$	$u^5$
$\sin(x^3)$	$x^3$	$\sin u$
$e^{\cos x}$	$\cos x$	$e^u$
$\sqrt{4 - x^2}$	$4 - x^2$	$\sqrt{u}$

A quick way to find the outer function: ask yourself, "What is the last thing I would do if I evaluated this on a calculator?" For $(3x^2 + 1)^5$ you would compute $3x^2 + 1$ first and raise it to the 5th power last, so the outer function is the 5th power.

Why rates multiply

Think of the composite as a chain of machines. Suppose $u = g(x)$ and $y = f(u)$ . If $u$ changes 3 times as fast as $x$ , and $y$ changes 2 times as fast as $u$ , then $y$ changes $2 \cdot 3 = 6$ times as fast as $x$ . Rates of change along a chain multiply.

In Leibniz notation this looks almost like canceling fractions:

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

(They are not really fractions, but the notation is designed so that this memory aid works.)

The chain rule

If $h(x) = f(g(x))$ , then

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

In words: differentiate the outer function, leaving the inside alone, then multiply by the derivative of the inside.

Notice what "leaving the inside alone" means. In $f'(g(x))$ , the derivative $f'$ is evaluated at $g(x)$ , not at $x$ . That detail matters a lot when you work from a table of values.

The graph below shows the effect of a simple inner function. The curve $y = \sin(2x)$ runs through a full wave twice as fast as $y = \sin x$ , so its slopes are twice as steep: $\dfrac{d}{dx}\sin(2x) = \cos(2x) \cdot 2$ .

y = sin x and y = sin(2x). The inner function 2x speeds everything up by a factor of 2, and the chain rule multiplies the slope by 2.Open in grapher →

Using the chain rule

Worked example: A power of a polynomial

Differentiate $y = (3x^2 + 1)^5$ .

Solution. The outer function is $u^5$ with derivative $5u^4$ . The inner function is $u = 3x^2 + 1$ with derivative $6x$ .

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

Expanding $(3x^2+1)^5$ first would also work, but it would take far longer and invite arithmetic mistakes.

Worked example: Trig, exponential and log outer functions

Differentiate each function.

$y = \sin(x^3)$
$y = e^{\cos x}$
$y = \ln(x^2 + 4)$

Solutions.

Outer $\sin u \to \cos u$ ; inner $x^3 \to 3x^2$ . So $y' = \cos(x^3) \cdot 3x^2 = 3x^2\cos(x^3)$ .
Outer $e^u \to e^u$ ; inner $\cos x \to -\sin x$ . So $y' = e^{\cos x} \cdot (-\sin x) = -\sin x \, e^{\cos x}$ .
Outer $\ln u \to \dfrac{1}{u}$ ; inner $x^2 + 4 \to 2x$ . So $y' = \dfrac{1}{x^2 + 4} \cdot 2x = \dfrac{2x}{x^2 + 4}$ .

The pattern in part 3 is worth remembering: $\dfrac{d}{dx}\ln(g(x)) = \dfrac{g'(x)}{g(x)}$ .

Worked example: More than two layers

Differentiate $y = \cos^3(2x)$ .

Solution. First rewrite so the layers are visible: $y = \big(\cos(2x)\big)^3$ . There are three layers: cube, then cosine, then $2x$ . Peel them from the outside in, multiplying as you go.

\begin{aligned} \frac{dy}{dx} &= 3\big(\cos(2x)\big)^2 \cdot \frac{d}{dx}\cos(2x) \\ &= 3\cos^2(2x) \cdot \big(-\sin(2x)\big) \cdot 2 \\ &= -6\cos^2(2x)\sin(2x). \end{aligned}

Worked example: Working from a table

The table gives values of differentiable functions $f$ and $g$ . Let $h(x) = f(g(x))$ . Find $h'(1)$ .

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

Solution. By the chain rule, $h'(1) = f'(g(1)) \cdot g'(1)$ . From the table, $g(1) = 3$ , so we need $f'(3) = 7$ , not $f'(1)$ . Then

h'(1) = f'(3) \cdot g'(1) = 7 \cdot (-2) = -14.

Common mistake

The two most common chain rule mistakes:

Forgetting the inner derivative. $\dfrac{d}{dx}\sin(5x)$ is $5\cos(5x)$ , not $\cos(5x)$ .
Evaluating $f'$ at the wrong input. In a table problem, $h'(a) = f'(g(a))\,g'(a)$ . Look up $g(a)$ first, then find $f'$ at that value.

Tip

When an answer has no inner derivative factor, check whether the inside was just $x$ . The chain rule is always in play; when the inner function is $x$ , its derivative is 1 and the factor is invisible.

On the AP exam the chain rule rarely appears alone. It sits inside product rule, quotient rule, implicit differentiation and related rates problems, so fluency here pays off for the rest of the year.