The Chain Rule

You already know how to differentiate $x^4$, $\sin x$, $e^x$, and $\ln x$. But what about $(3x^2 - 5)^4$, $\sin(5x^2)$, or $e^{-3x}$? These are composite functions: one function sitting inside another. The chain rule tells you how to differentiate them, and it’s the rule you’ll use more than any other in calculus. All trig on this page is in radians, as in all of AP Calculus.

Key ideas

Inside and outside

A composite function $f(g(x))$ has an outside function $f$ and an inside function $g$. To spot them, ask: “If I were evaluating this on a calculator, what would I do last?” The last step is the outside.

Function	Inside $g(x)$	Outside $f(u)$
$(3x^2 - 5)^4$	$3x^2 - 5$	$u^4$
$\sin(5x^2)$	$5x^2$	$\sin u$
$e^{-3x}$	$-3x$	$e^u$
$\ln(x^2 + 4)$	$x^2 + 4$	$\ln u$
$\cos^3 x = (\cos x)^3$	$\cos x$	$u^3$

The chain rule

$$\frac{d}{dx}\Big[f\big(g(x)\big)\Big] = f'\big(g(x)\big) \cdot g'(x)$$

In words: differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.

Leibniz form

If $y = f(u)$ and $u = g(x)$, then

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

This form shows why the rule works: if $y$ changes $3$ times as fast as $u$, and $u$ changes $2$ times as fast as $x$, then $y$ changes $3 \times 2 = 6$ times as fast as $x$. Rates of change multiply along the chain. (The $du$‘s look like they cancel. They aren’t really fractions, but it’s a handy memory aid.)

Common patterns

Each basic rule from derivatives of trig, exponential, and log functions gets a “times $u'$” at the end, where $u$ is the inside function:

Function	Derivative
$u^n$	$n u^{n-1} \cdot u'$
$\sin u$	$\cos u \cdot u'$
$\cos u$	$-\sin u \cdot u'$
$\tan u$	$\sec^2 u \cdot u'$
$e^u$	$e^u \cdot u'$
$a^u$	$a^u \ln a \cdot u'$
$\ln u$	$\dfrac{u'}{u}$

For more than two layers, keep going: work from the outside in, multiplying by the derivative of each layer. For example, $\sin^3(2x)$ has three layers (cube, sine, $2x$):

$$\frac{d}{dx}\big[\sin(2x)\big]^3 = 3\sin^2(2x) \cdot \cos(2x) \cdot 2$$

Using a table of values

If $h(x) = f(g(x))$, then $h'(a) = f'(g(a)) \cdot g'(a)$. With a table, work in this order:

1. Find $g(a)$ in the table.
2. Look up $f'$ at that value, $g(a)$, not at $a$.
3. Multiply by $g'(a)$.

Choosing the right procedure

Real functions often need several rules at once (CED 3.5). Look at the overall structure first:

• Is the whole thing a product or quotient of two pieces? Start with the product rule or quotient rule, and use the chain rule on any piece that’s composite.
• Is the whole thing one function of another? Start with the chain rule.
• Rewrite first when it makes life easier: $\sqrt{x^2 + 9} = (x^2 + 9)^{1/2}$, and $\dfrac{4}{(x^2 + 1)^3} = 4(x^2 + 1)^{-3}$ needs only the chain rule, not the quotient rule.

On the AP exam, these are mostly no-calculator skills, so practise until they’re automatic.

Worked examples

Example 1: A power of a polynomial

Differentiate $y = (3x^2 - 5)^4$.

Solution. The outside is $u^4$ and the inside is $u = 3x^2 - 5$, with $u' = 6x$.

$$\frac{dy}{dx} = 4(3x^2 - 5)^3 \cdot 6x = 24x(3x^2 - 5)^3$$

Don’t expand $(3x^2 - 5)^4$ first. It works, but it’s slow and easy to get wrong.

Example 2: Trig, exponential, and log compositions

Differentiate each function.

(a) $\sin(5x^2)$ (b) $e^{-3x}$ (c) $\ln(x^2 + 4)$ (d) $\cos^3 x$

Solution.

(a) Outside $\sin u$, inside $5x^2$:

$$\frac{d}{dx}\sin(5x^2) = \cos(5x^2) \cdot 10x = 10x\cos(5x^2)$$

(b) Outside $e^u$, inside $-3x$:

$$\frac{d}{dx}e^{-3x} = e^{-3x} \cdot (-3) = -3e^{-3x}$$

(c) Outside $\ln u$, inside $x^2 + 4$:

$$\frac{d}{dx}\ln(x^2 + 4) = \frac{1}{x^2 + 4} \cdot 2x = \frac{2x}{x^2 + 4}$$

(d) Rewrite as $(\cos x)^3$. Outside $u^3$, inside $\cos x$:

$$\frac{d}{dx}(\cos x)^3 = 3(\cos x)^2 \cdot (-\sin x) = -3\cos^2 x \sin x$$

Example 3: Using a table

The table gives values of $f$, $g$, and their derivatives.

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

(a) Let $h(x) = f(g(x))$. Find $h'(1)$.

(b) Let $k(x) = \big(f(x)\big)^2$. Find $k'(3)$.

Solution.

(a) $h'(1) = f'(g(1)) \cdot g'(1)$. From the table, $g(1) = 3$, so we need $f'(3) = 6$:

$$h'(1) = f'(3) \cdot g'(1) = 6 \cdot 5 = 30$$

(b) The outside is $u^2$ and the inside is $f(x)$, so $k'(x) = 2f(x) \cdot f'(x)$:

$$k'(3) = 2f(3) \cdot f'(3) = 2(7)(6) = 84$$

Example 4: Product rule and chain rule together

Find the equation of the tangent line to $y = x^2\sqrt{1 + 4x}$ at $x = 2$.

Solution. Overall, this is a product of $x^2$ and $(1 + 4x)^{1/2}$, so start with the product rule. The second factor needs the chain rule:

$$\frac{d}{dx}(1 + 4x)^{1/2} = \frac{1}{2}(1 + 4x)^{-1/2} \cdot 4 = \frac{2}{\sqrt{1 + 4x}}$$

So

$$\frac{dy}{dx} = 2x\sqrt{1 + 4x} + x^2 \cdot \frac{2}{\sqrt{1 + 4x}}$$

At $x = 2$: $\sqrt{1 + 8} = 3$, so

$$y = 4(3) = 12, \qquad \frac{dy}{dx} = 2(2)(3) + \frac{4(2)}{3} = 12 + \frac{8}{3} = \frac{44}{3}$$

The tangent line is

$$y - 12 = \frac{44}{3}(x - 2)$$

Common mistakes

Forgetting to multiply by the inside derivative. $\dfrac{d}{dx}\sin(3x)$ is $3\cos(3x)$, not $\cos(3x)$. Even when the inside is something simple like $3x$ or $-x$, its derivative still counts.

Changing the inside when differentiating the outside. $\dfrac{d}{dx}\cos(x^2) = -\sin(x^2) \cdot 2x$. A common wrong answer is $-\sin(2x)$, which differentiates the inside inside the cosine. The inside stays exactly as it was; its derivative goes out front.

Using the wrong values from a table. For $h(x) = f(g(x))$, $h'(a) = f'(g(a)) \cdot g'(a)$. Students often write $f'(a) \cdot g'(a)$ or $f'(g'(a))$. Always find $g(a)$ first, then look up $f'$ there.

Confusing sin²x with sin(x²). $\sin^2 x = (\sin x)^2$ has outside $u^2$, so its derivative is $2\sin x\cos x$. But $\sin(x^2)$ has outside $\sin u$, so its derivative is $2x\cos(x^2)$.

Dropping the inside derivative with ln. $\dfrac{d}{dx}\ln(x^2 + 1) = \dfrac{2x}{x^2 + 1}$, not $\dfrac{1}{x^2 + 1}$. Remember the pattern “derivative of the inside over the inside”.

Using degrees. Calculus derivative rules like $\dfrac{d}{dx}\sin x = \cos x$ only work in radians. Make sure your calculator is in radian mode on the AP exam.

Practice

1. (Warm-up) Differentiate $y = (2x + 7)^5$.

Solution

Outside $u^5$, inside $2x + 7$ with derivative $2$:

$$\frac{dy}{dx} = 5(2x + 7)^4 \cdot 2 = 10(2x + 7)^4$$

2. (Warm-up) Differentiate each function.

• (a) $e^{4x}$
• (b) $\sqrt{x^2 + 9}$

Solution

(a) $\dfrac{d}{dx}e^{4x} = e^{4x} \cdot 4 = 4e^{4x}$

(b) Rewrite as $(x^2 + 9)^{1/2}$:

$$\frac{d}{dx}(x^2 + 9)^{1/2} = \frac{1}{2}(x^2 + 9)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 9}}$$

3. (Core) Differentiate $y = \ln(\cos x)$ and simplify.

Solution

Outside $\ln u$, inside $\cos x$:

$$\frac{dy}{dx} = \frac{-\sin x}{\cos x} = -\tan x$$

4. (Core) Let $y = u^3 - 2u$ and $u = x^2 + 1$. Use $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ to find $\dfrac{dy}{dx}$ when $x = 1$.

Solution

$\dfrac{dy}{du} = 3u^2 - 2$ and $\dfrac{du}{dx} = 2x$.

When $x = 1$, $u = 1^2 + 1 = 2$, so

$$\frac{dy}{dx} = (3(2)^2 - 2)(2 \cdot 1) = 10 \cdot 2 = 20$$

5. (Core) Use the table.

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$0$	$2$	$3$	$5$	$-1$
$2$	$0$	$-4$	$1$	$6$

• (a) If $h(x) = g(f(x))$, find $h'(0)$.
• (b) If $k(x) = f(f(x))$, find $k'(2)$.
• (c) If $m(x) = e^{g(x)}$, find $m'(2)$.

Solution

(a) $h'(0) = g'(f(0)) \cdot f'(0) = g'(2) \cdot 3 = 6 \cdot 3 = 18$

(b) $k'(2) = f'(f(2)) \cdot f'(2) = f'(0) \cdot (-4) = 3(-4) = -12$

(c) $m'(x) = e^{g(x)} \cdot g'(x)$, so $m'(2) = e^{g(2)} \cdot g'(2) = e^1 \cdot 6 = 6e$

6. (Core) Differentiate $y = \dfrac{4}{(x^2 + 1)^3}$ without using the quotient rule.

Solution

Rewrite as $y = 4(x^2 + 1)^{-3}$:

$$\frac{dy}{dx} = 4 \cdot (-3)(x^2 + 1)^{-4} \cdot 2x = -\frac{24x}{(x^2 + 1)^4}$$

7. (Core) The depth of water at a dock is $D(t) = 3 + 1.5\sin\left(\dfrac{\pi t}{6}\right)$ metres, $t$ hours after midnight. How fast is the depth changing at $t = 2$? Give an exact answer and a decimal to three places, with units.

Solution

$$D'(t) = 1.5\cos\left(\frac{\pi t}{6}\right) \cdot \frac{\pi}{6} = \frac{\pi}{4}\cos\left(\frac{\pi t}{6}\right)$$$$D'(2) = \frac{\pi}{4}\cos\left(\frac{\pi}{3}\right) = \frac{\pi}{4} \cdot \frac{1}{2} = \frac{\pi}{8} \approx 0.393$$

The depth is increasing at about $0.393$ metres per hour.

8. (Challenge) Let $y = \sin^3(2x)$. Find $\dfrac{dy}{dx}$, then evaluate it at $x = \dfrac{\pi}{12}$.

Solution

Three layers: the cube, the sine, and $2x$.

$$\frac{dy}{dx} = 3\sin^2(2x) \cdot \cos(2x) \cdot 2 = 6\sin^2(2x)\cos(2x)$$

At $x = \dfrac{\pi}{12}$, $2x = \dfrac{\pi}{6}$, with $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$ and $\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$:

$$\frac{dy}{dx} = 6 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}$$

9. (Challenge) Find all points where the graph of $y = xe^{-x^2}$ has a horizontal tangent line.

Solution

Product rule, with the chain rule on $e^{-x^2}$:

$$\frac{dy}{dx} = 1 \cdot e^{-x^2} + x \cdot e^{-x^2}(-2x) = e^{-x^2}(1 - 2x^2)$$

Since $e^{-x^2} \gt 0$ always, $\dfrac{dy}{dx} = 0$ only when $1 - 2x^2 = 0$, so $x = \pm\dfrac{1}{\sqrt{2}}$.

The $y$-values are $y = \pm\dfrac{1}{\sqrt{2}}e^{-1/2}$, so the points are

$$\left(\frac{1}{\sqrt{2}},\ \frac{1}{\sqrt{2}}e^{-1/2}\right) \quad\text{and}\quad \left(-\frac{1}{\sqrt{2}},\ -\frac{1}{\sqrt{2}}e^{-1/2}\right)$$

That’s about $(0.707, 0.429)$ and $(-0.707, -0.429)$.