Chain rule | AP Calc AB Advanced differentiation

What you’ll learn

How to differentiate composite functions with the chain rule
Applying the chain rule with logarithmic, exponential, trig, and power functions
Combining the chain rule with product and quotient rules

To find the derivative of a function like $h (x) = (2 x + 1)^{3}$ , expanding and then applying the power rule is correct but inefficient.

Instead, a faster method called the chain rule is used to differentiate composite functions - functions built by nesting one function into another.

Chain rule:

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

In other words,

Differentiate the outer function, keeping the inner function intact.
Multiply it by the derivative of the inner function.

Another common way to write the chain rule uses Leibniz notation:

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

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Examples

Differentiate

$y = (2 x + 1)^{3}$

Solution

(spoiler)

$y$ is a composite function, so we identify the inner and outer parts. Below, the inner function is boxed:

$outer (2 x + 1 inner)^{3}$

For clarity,

Outer function: $f (u) = u^{3}$
Inner function: $g (x) = 2 x + 1$

Apply the Chain Rule: differentiate the outer function (using the power rule) while leaving the inner function untouched, then multiply by the derivative of the inner function.

$y^{'} = 3 (2 x + 1)^{2} derivative of outer \cdot \frac{d}{d x} (2 x + 1) derivative of inner = 3 (2 x + 1)^{2} \cdot (2) = 6 (2 x + 1)^{2}$

Let’s try one with the natural logarithm. Recall from section 2.5 that $\frac{d}{d x} [ln (x)] = \frac{1}{x}$ .

Differentiate

$h (x) = ln (3 x^{2} + x)$

Solution

(spoiler)

$h (x)$ is a composite function $f (g (x))$ with:

$outer ln (3 x^{2} + x inner)$

For clarity,

Outer function: $f (u) = ln (u)$
Inner function: $g (x) = 3 x^{2} + x$

Applying the chain rule, differentiate the outer function while keeping the inner function intact, then multiply by the derivative of the inner function:

$h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = \frac{1}{3 x ^{2} + x} \cdot \frac{d}{d x} (3 x^{2} + x) = \frac{1}{3 x ^{2} + x} \cdot (6 x + 1) = \frac{6 x + 1}{3 x ^{2} + x}$

Similarly, recall that for exponential functions,

$\frac{d}{d x} (e^{x}) = e^{x}$

and

$\frac{d}{d x} (b^{x}) = b^{x} ln (b)$

Differentiate:

a) $a (x) = e^{- 2 x}$

b) $b (x) = 2^{3 x + 1}$

Solutions

a) $a (x) = e^{- 2 x}$

(spoiler)

$a (x)$ is a composite function, where

Outer function: $f (u) = e^{u}$
Inner function: $g (x) = - 2 x$

Applying the chain rule,

$a^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = e^{- 2 x} \cdot \frac{d}{d x} (- 2 x) = e^{- 2 x} \cdot (- 2) = - 2 e^{- 2 x}$

b) $b (x) = 2^{3 x + 1}$

(spoiler)

$b (x)$ consists of

Outer function: $f (u) = 2^{u}$
Inner function: $g (x) = 3 x + 1$

$b^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = 2^{3 x + 1} ln (2) \cdot \frac{d}{d x} (3 x + 1) = 2^{3 x + 1} ln (2) \cdot (3) = 3 ln (2) \cdot 2^{3 x + 1}$

Combining rules

The chain rule often appears together with other derivative rules. The main thing to watch is order: decide which overall structure you have first (product, quotient, or composition), then apply the needed rules inside that structure.

For example, the following problem uses the product rule first, and then uses the chain rule while differentiating one of the factors.

Differentiate

$h (x) = 3 x (x^{2} - 1)^{3}$

Solution

(spoiler)

$h (x)$ is a product of two functions:

$f (x) = 3 x$
$g (x) = (x^{2} - 1)^{3}$

Notice that $g (x)$ is composite, so the chain rule is needed to find $g^{'} (x)$ .

Function	Expression
$f (x)$	$3 x$
$f^{'} (x)$	$3$
$g (x)$	$(x^{2} - 1)^{3}$
$g^{'} (x)$	$3 (x^{2} - 1)^{2} \cdot 2 x$

Putting these into the product rule:

$h^{'} (x) = f (x) g^{'} (x) + f^{'} (x) g (x) = 3 x \cdot [3 (x^{2} - 1)^{2} \cdot 2 x] + 3 \cdot (x^{2} - 1)^{3} = 18 x^{2} (x^{2} - 1)^{2} + 3 (x^{2} - 1)^{3}$

To simplify, factor out the common factors $3$ and $(x^{2} - 1)^{2}$ :

$= 3 (x^{2} - 1)^{2} [6 x^{2} + (x^{2} - 1)] = 3 (x^{2} - 1)^{2} (7 x^{2} - 1)$

Contrast that with the next problem, which uses the power rule first, and then the product rule when applying the chain rule and differentiating the inner function.

Differentiate

$h (x) = (e^{x} sin x)^{2}$

Solution

(spoiler)

$h (x)$ is a composite function, where:

Outer function: $f (u) = u^{2}$
Inner function: $g (x) = e^{x} sin x$

Applying the chain rule,

$h^{'} (x) = 2 (e^{x} sin x) \cdot \frac{d}{d x} (e^{x} sin x)$

To differentiate $e^{x} sin x)$ , apply the product rule

$\frac{d}{d x} (e^{x} sin x) = e^{x} cos x + e^{x} sin x = e^{x} (cos x + sin x)$

Then

$h^{'} (x) = 2 e^{x} sin x \cdot e^{x} (cos x + sin x) = 2 e^{2 x} sin x (cos x + sin x)$

Differentiate

$h (x) = \frac{x}{2 ( x ^{2} + 1 ) ^{2}}$

Solution

(spoiler)

For easier algebra, pull out the constant multiple $\frac{1}{2}$ :

$h (x) = \frac{1}{2} \cdot \frac{x}{( x ^{2} + 1 ) ^{2}}$

Now $h (x)$ is a quotient of two functions:

$f (x) = x$
$g (x) = (x^{2} + 1)^{2}$

Use the quotient rule. Note that $g (x)$ is composite, so the chain rule is needed for $g^{'} (x)$ .

Function	Expression
$f (x)$	$x$
$f^{'} (x)$	$1$
$g (x)$	$(x^{2} + 1)^{2}$
$g^{'} (x)$	$4 x (x^{2} + 1)$

Putting the pieces into the quotient rule:

$h^{'} (x) = \frac{1}{2} \cdot \frac{g ( x ) f ^{'} ( x ) - f ( x ) g ^{'} ( x )}{( g ( x ) ) ^{2}} = \frac{1}{2} \cdot \frac{( x ^{2} + 1 ) ^{2} ( 1 ) - x \cdot 4 x ( x ^{2} + 1 )}{( x ^{2} + 1 ) ^{4}} = \frac{( x ^{2} + 1 ) ^{2} - 4 x ^{2} ( x ^{2} + 1 )}{2 ( x ^{2} + 1 ) ^{4}} = \frac{( x ^{2} + 1 ) ( x ^{2} + 1 - 4 x ^{2} )}{2 ( x ^{2} + 1 ) ^{4}} = \frac{1 - 3 x ^{2}}{2 ( x ^{2} + 1 ) ^{3}}$

You can also reach the same answer by rewriting

$h (x) = \frac{1}{2} x (x^{2} + 1)^{- 2}$

and applying the product rule while using the chain rule on $(x^{2} + 1)^{- 2}$ .

Using tables

Some questions ask you to apply derivative rules using only a table of values:

$f (x), g (x),$ and their derivatives have the following values shown in the table:

$x$ $f (x)$ $g (x)$ $f^{'} (x)$ $g^{'} (x)$

$1$ $3$ $2$ $- 1$ $5$

$2$ $0.2$ $0.5$ $5$ $3$

$3$ $- 1$ $1$ $4$ $- 3$

Find $h^{'} (1)$ given

$h (x) = f (g (x) - x^{2})$

$x$	$f (x)$	$g (x)$	$f^{'} (x)$	$g^{'} (x)$
$1$	$3$	$2$	$- 1$	$5$
$2$	$0.2$	$0.5$	$5$	$3$
$3$	$- 1$	$1$	$4$	$- 3$

Solution

(spoiler)

Because $h (x) = f (g (x) - x^{2})$ is a composite function, apply the chain rule:

$\frac{d}{d x} [f (g (x) - x^{2})]$

$= f^{'} (g (x) - x^{2}) \cdot (g^{'} (x) - 2 x)$

Now evaluate at $x = 1$ :

$h^{'} (1) = f^{'} (g (1) - 1^{2}) \cdot (g^{'} (1) - 2 (1)) = f^{'} (2 - 1) \cdot (5 - 2) = f^{'} (1) \cdot 3 = - 1 \cdot 3 = - 3$

Challenge problems

Differentiate

$h (t) = 3 2 π - π t$

Solution

(spoiler)

Rewrite the square root as a power:

$h (t) = 3 (2 π - π t)^{1/2}$

Then $h (t)$ consists of

Outer: $f (u) = 3 u^{1/2}$
Inner: $g (t) = 2 π - π t$

Applying the chain rule (remembering that $π$ is a constant),

$h^{'} (t) = f^{'} (g (t)) \cdot g^{'} (t) = \frac{3}{2} (2 π - π t)^{- 1/2} \cdot \frac{d}{d t} (2 π - π t) = \frac{3}{2} (2 π - π t)^{- 1/2} \cdot (- π) = - \frac{3 π}{2 2 π - π t}$

Differentiate

$h (x) = \frac{e ^{x^{2}}}{e ^{c o s x}}$

Solution

(spoiler)

Instead of using the quotient rule, rewrite the expression using $\frac{a ^{m}}{a ^{n}} = a^{m - n}$ :

$h (x) = e^{x^{2} - c o s x}$

Then $h (x)$ consists of

Outer: $f (u) = e^{u}$
Inner: $g (x) = x^{2} - cos x$

Applying the chain rule,

$h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = e^{x^{2} - c o s x} \cdot \frac{d}{d x} (x^{2} - cos x) = e^{x^{2} - c o s x} \cdot (2 x - (- sin x)) = (e^{x^{2} - c o s x}) (2 x + sin x)$

Differentiate

$A (x) = sin^{3} (x^{2})$

Solution

(spoiler)

First, rewrite as

$A (x) = [sin (x^{2})]^{3}$

This function has multiple layers of composition and requires two applications of the chain rule:

Outer: $f (u) = u^{3} ⟹ f^{'} (u) = 3 u^{2}$
Middle: $g (v) = sin (v) ⟹ g^{'} (v) = cos (v)$
Inner: $h (x) = x^{2} ⟹ h^{'} (x) = 2 x$

Apply the chain rule by “unfolding” each layer:

$A^{'} (x) = 3 [sin (x^{2})]^{2} \cdot \frac{d}{d x} [sin (x^{2})] = 3 [sin (x^{2})]^{2} \cdot cos (x^{2}) \cdot \frac{d}{d x} [x^{2}] = 3 [sin (x^{2})]^{2} \cdot cos (x^{2}) \cdot (2 x) = 6 x sin^{2} (x^{2}) cos (x^{2})$

What you’ll learn

How to differentiate composite functions with the chain rule
Applying the chain rule with logarithmic, exponential, trig, and power functions
Combining the chain rule with product and quotient rules

To find the derivative of a function like $h (x) = (2 x + 1)^{3}$ , expanding and then applying the power rule is correct but inefficient.

Instead, a faster method called the chain rule is used to differentiate composite functions - functions built by nesting one function into another.

Chain rule:

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

In other words,

Differentiate the outer function, keeping the inner function intact.
Multiply it by the derivative of the inner function.

Another common way to write the chain rule uses Leibniz notation:

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

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$

Examples

Differentiate

$y = (2 x + 1)^{3}$

Solution

(spoiler)

$y$ is a composite function, so we identify the inner and outer parts. Below, the inner function is boxed:

$outer (2 x + 1 inner)^{3}$

For clarity,

Outer function: $f (u) = u^{3}$
Inner function: $g (x) = 2 x + 1$

Apply the Chain Rule: differentiate the outer function (using the power rule) while leaving the inner function untouched, then multiply by the derivative of the inner function.

$y^{'} = 3 (2 x + 1)^{2} derivative of outer \cdot \frac{d}{d x} (2 x + 1) derivative of inner = 3 (2 x + 1)^{2} \cdot (2) = 6 (2 x + 1)^{2}$

Let’s try one with the natural logarithm. Recall from section 2.5 that $\frac{d}{d x} [ln (x)] = \frac{1}{x}$ .

Differentiate

$h (x) = ln (3 x^{2} + x)$

Solution

(spoiler)

$h (x)$ is a composite function $f (g (x))$ with:

$outer ln (3 x^{2} + x inner)$

For clarity,

Outer function: $f (u) = ln (u)$
Inner function: $g (x) = 3 x^{2} + x$

Applying the chain rule, differentiate the outer function while keeping the inner function intact, then multiply by the derivative of the inner function:

$h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = \frac{1}{3 x ^{2} + x} \cdot \frac{d}{d x} (3 x^{2} + x) = \frac{1}{3 x ^{2} + x} \cdot (6 x + 1) = \frac{6 x + 1}{3 x ^{2} + x}$

Similarly, recall that for exponential functions,

$\frac{d}{d x} (e^{x}) = e^{x}$

and

$\frac{d}{d x} (b^{x}) = b^{x} ln (b)$

Differentiate:

a) $a (x) = e^{- 2 x}$

b) $b (x) = 2^{3 x + 1}$

Solutions

a) $a (x) = e^{- 2 x}$

(spoiler)

$a (x)$ is a composite function, where

Outer function: $f (u) = e^{u}$
Inner function: $g (x) = - 2 x$

Applying the chain rule,

$a^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = e^{- 2 x} \cdot \frac{d}{d x} (- 2 x) = e^{- 2 x} \cdot (- 2) = - 2 e^{- 2 x}$

b) $b (x) = 2^{3 x + 1}$

(spoiler)

$b (x)$ consists of

Outer function: $f (u) = 2^{u}$
Inner function: $g (x) = 3 x + 1$

$b^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = 2^{3 x + 1} ln (2) \cdot \frac{d}{d x} (3 x + 1) = 2^{3 x + 1} ln (2) \cdot (3) = 3 ln (2) \cdot 2^{3 x + 1}$

Combining rules

For example, the following problem uses the product rule first, and then uses the chain rule while differentiating one of the factors.

Differentiate

$h (x) = 3 x (x^{2} - 1)^{3}$

Solution

(spoiler)

$h (x)$ is a product of two functions:

$f (x) = 3 x$
$g (x) = (x^{2} - 1)^{3}$

Notice that $g (x)$ is composite, so the chain rule is needed to find $g^{'} (x)$ .

Function	Expression
$f (x)$	$3 x$
$f^{'} (x)$	$3$
$g (x)$	$(x^{2} - 1)^{3}$
$g^{'} (x)$	$3 (x^{2} - 1)^{2} \cdot 2 x$

Putting these into the product rule:

$h^{'} (x) = f (x) g^{'} (x) + f^{'} (x) g (x) = 3 x \cdot [3 (x^{2} - 1)^{2} \cdot 2 x] + 3 \cdot (x^{2} - 1)^{3} = 18 x^{2} (x^{2} - 1)^{2} + 3 (x^{2} - 1)^{3}$

To simplify, factor out the common factors $3$ and $(x^{2} - 1)^{2}$ :

$= 3 (x^{2} - 1)^{2} [6 x^{2} + (x^{2} - 1)] = 3 (x^{2} - 1)^{2} (7 x^{2} - 1)$

Contrast that with the next problem, which uses the power rule first, and then the product rule when applying the chain rule and differentiating the inner function.

Differentiate

$h (x) = (e^{x} sin x)^{2}$

Solution

(spoiler)

$h (x)$ is a composite function, where:

Outer function: $f (u) = u^{2}$
Inner function: $g (x) = e^{x} sin x$

Applying the chain rule,

$h^{'} (x) = 2 (e^{x} sin x) \cdot \frac{d}{d x} (e^{x} sin x)$

To differentiate $e^{x} sin x)$ , apply the product rule

$\frac{d}{d x} (e^{x} sin x) = e^{x} cos x + e^{x} sin x = e^{x} (cos x + sin x)$

Then

$h^{'} (x) = 2 e^{x} sin x \cdot e^{x} (cos x + sin x) = 2 e^{2 x} sin x (cos x + sin x)$

Differentiate

$h (x) = \frac{x}{2 ( x ^{2} + 1 ) ^{2}}$

Solution

(spoiler)

For easier algebra, pull out the constant multiple $\frac{1}{2}$ :

$h (x) = \frac{1}{2} \cdot \frac{x}{( x ^{2} + 1 ) ^{2}}$

Now $h (x)$ is a quotient of two functions:

$f (x) = x$
$g (x) = (x^{2} + 1)^{2}$

Use the quotient rule. Note that $g (x)$ is composite, so the chain rule is needed for $g^{'} (x)$ .

Function	Expression
$f (x)$	$x$
$f^{'} (x)$	$1$
$g (x)$	$(x^{2} + 1)^{2}$
$g^{'} (x)$	$4 x (x^{2} + 1)$

Putting the pieces into the quotient rule:

You can also reach the same answer by rewriting

$h (x) = \frac{1}{2} x (x^{2} + 1)^{- 2}$

and applying the product rule while using the chain rule on $(x^{2} + 1)^{- 2}$ .

Using tables

Some questions ask you to apply derivative rules using only a table of values:

$f (x), g (x),$ and their derivatives have the following values shown in the table:

$x$ $f (x)$ $g (x)$ $f^{'} (x)$ $g^{'} (x)$

$1$ $3$ $2$ $- 1$ $5$

$2$ $0.2$ $0.5$ $5$ $3$

$3$ $- 1$ $1$ $4$ $- 3$

Find $h^{'} (1)$ given

$h (x) = f (g (x) - x^{2})$

$x$	$f (x)$	$g (x)$	$f^{'} (x)$	$g^{'} (x)$
$1$	$3$	$2$	$- 1$	$5$
$2$	$0.2$	$0.5$	$5$	$3$
$3$	$- 1$	$1$	$4$	$- 3$

Solution

(spoiler)

Because $h (x) = f (g (x) - x^{2})$ is a composite function, apply the chain rule:

$\frac{d}{d x} [f (g (x) - x^{2})]$

$= f^{'} (g (x) - x^{2}) \cdot (g^{'} (x) - 2 x)$

Now evaluate at $x = 1$ :

$h^{'} (1) = f^{'} (g (1) - 1^{2}) \cdot (g^{'} (1) - 2 (1)) = f^{'} (2 - 1) \cdot (5 - 2) = f^{'} (1) \cdot 3 = - 1 \cdot 3 = - 3$

Challenge problems

Differentiate

$h (t) = 3 2 π - π t$

Solution

(spoiler)

Rewrite the square root as a power:

$h (t) = 3 (2 π - π t)^{1/2}$

Then $h (t)$ consists of

Outer: $f (u) = 3 u^{1/2}$
Inner: $g (t) = 2 π - π t$

Applying the chain rule (remembering that $π$ is a constant),

$h^{'} (t) = f^{'} (g (t)) \cdot g^{'} (t) = \frac{3}{2} (2 π - π t)^{- 1/2} \cdot \frac{d}{d t} (2 π - π t) = \frac{3}{2} (2 π - π t)^{- 1/2} \cdot (- π) = - \frac{3 π}{2 2 π - π t}$

Differentiate

$h (x) = \frac{e ^{x^{2}}}{e ^{c o s x}}$

Solution

(spoiler)

Instead of using the quotient rule, rewrite the expression using $\frac{a ^{m}}{a ^{n}} = a^{m - n}$ :

$h (x) = e^{x^{2} - c o s x}$

Then $h (x)$ consists of

Outer: $f (u) = e^{u}$
Inner: $g (x) = x^{2} - cos x$

Applying the chain rule,

$h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x) = e^{x^{2} - c o s x} \cdot \frac{d}{d x} (x^{2} - cos x) = e^{x^{2} - c o s x} \cdot (2 x - (- sin x)) = (e^{x^{2} - c o s x}) (2 x + sin x)$

Differentiate

$A (x) = sin^{3} (x^{2})$

Solution

(spoiler)

First, rewrite as

$A (x) = [sin (x^{2})]^{3}$

This function has multiple layers of composition and requires two applications of the chain rule:

Outer: $f (u) = u^{3} ⟹ f^{'} (u) = 3 u^{2}$
Middle: $g (v) = sin (v) ⟹ g^{'} (v) = cos (v)$
Inner: $h (x) = x^{2} ⟹ h^{'} (x) = 2 x$

Apply the chain rule by “unfolding” each layer: