Motivating Questions

What is a composite function and how do we recognize its structure algebraically?
Given a composite function $C(x) = f(g(x))$ that is built from differentiable functions $f$ and $g\text{,}$ how do we compute $C'(x)$ in terms of $f\text{,}$ $g\text{,}$ $f'\text{,}$ and $g'\text{?}$ What is the statement of the Chain Rule?

In addition to learning how to differentiate a variety of basic functions, we have also been developing our ability to use rules to differentiate certain algebraic combinations of them.

Example $2.5.1$

State the rule(s) to find the derivative of each of the following combinations of $f(x) = \sin(x)$ and $g(x) = x^2\text{:}$

$s(x) = 3x^2 - 5\sin(x)\text{,}$

$p(x) = x^2 \sin(x), \text{and}$

$q(x) = \dfrac{\sin(x)}{x^2}\text{.}$

Answer: Finding $s'$ uses the sum and constant multiple rules, because $s(x) = 3g(x) - 5f(x)\text{.}$ Determining $p'$ requires the product rule, because $p(x) = g(x) \cdot f(x)\text{.}$ To calculate $q'$ we use the quotient rule, because $q(x) =\frac{f(x)}{g(x)}\text{.}$

There is one more natural way to combine basic functions algebraically, and that is by composing them. For instance, let's consider the function

C(x) = \sin(x^2)\text{,}

and observe that any input $x$ passes through a chain of functions. In the process that defines the function $C(x)\text{,}$ $x$ is first squared, and then the sine of the result is taken. Using an arrow diagram,

x \longrightarrow x^2 \longrightarrow \sin(x^2)\text{.}

In terms of the elementary functions $f$ and $g\text{,}$ we observe that $x$ is the input for the function $g\text{,}$ and the result is used as the input for $f\text{.}$ We write

C(x) = f(g(x)) = \sin(x^2)

and say that $C$ is the composition of $f$ and $g\text{.}$ We will refer to $g\text{,}$ the function that is first applied to $x\text{,}$ as the inner function, while $f\text{,}$ the function that is applied to the result, is the outer function.

Given a composite function $C(x) = f(g(x))$ that is built from differentiable functions $f$ and $g\text{,}$ how do we compute $C'(x)$ in terms of $f\text{,}$ $g\text{,}$ $f'\text{,}$ and $g'\text{?}$ In the same way that the rate of change of a product of two functions, $p(x) = f(x) \cdot g(x)\text{,}$ depends on the behavior of both $f$ and $g\text{,}$ it makes sense intuitively that the rate of change of a composite function $C(x) = f(g(x))$ will also depend on some combination of $f$ and $g$ and their derivatives. The rule that describes how to compute $C'$ in terms of $f$ and $g$ and their derivatives is called the chain rule.

But before we can learn what the chain rule says and why it works, we first need to be comfortable decomposing composite functions so that we can correctly identify the inner and outer functions, as we did in the example above with $C(x) = \sin(x^2)\text{.}$

Preview Activity $2.5.1$

For each function given below, identify its fundamental algebraic structure. In particular, is the given function a sum, product, quotient, or composition of basic functions? If the function is a composition of basic functions, state a formula for the inner function $g$ and the outer function $f$ so that the overall composite function can be written in the form $f(g(x))\text{.}$ If the function is a sum, product, or quotient of basic functions, use the appropriate rule to determine its derivative.

$\displaystyle h(x) = \tan(2^x)$
$\displaystyle p(x) = 2^x \tan(x)$
$\displaystyle r(x) = (\tan(x))^2$
$\displaystyle m(x) = e^{\tan(x)}$
$\displaystyle w(x) = \sqrt{x} + \tan(x)$
$\displaystyle z(x) = \sqrt{\tan(x)}$

Often a composite function cannot be written in an alternate algebraic form. For instance, the function $C(x) = \sin(x^2)$ cannot be expanded or otherwise rewritten, so it presents no alternate approaches to taking the derivative. But some composite functions can be expanded or simplified, and these provide a way to explore how the chain rule works.

Example $2.5.2$

Let $f(x) = -4x + 7$ and $g(x) = 3x - 5\text{.}$ Determine a formula for $C(x) = f(g(x))$ and compute $C'(x)\text{.}$ How is $C'$ related to $f$ and $g$ and their derivatives?

Answer

By the rules given for $f$ and $g\text{,}$

$\begin{align*} C(x) =\mathstrut & f(g(x))\\[4pt] =\mathstrut & f(3x-5)\\[4pt] =\mathstrut & -4(3x-5) + 7\\[4pt] =\mathstrut & -12x + 20 + 7\\[4pt] =\mathstrut & -12x + 27\text{.} \end{align*}$

Thus, $C'(x) = -12\text{.}$ Noting that $f'(x) = -4$ and $g'(x) = 3\text{,}$ we observe that $C'$ appears to be the product of $f'$ and $g'\text{.}$

It may seem that Example $2.5.2$ is too elementary to illustrate how to differentiate a composite function. Linear functions are the simplest of all functions, and composing linear functions yields another linear function. While this example does not illustrate the full complexity of a composition of nonlinear functions, at the same time we remember that any differentiable function is locally linear, and thus any function with a derivative behaves like a line when viewed up close. The fact that the derivatives of the linear functions $f$ and $g$ are multiplied to find the derivative of their composition turns out to be a key insight.

We now consider a composition involving a nonlinear function.

Example $2.5.3$

Let $C(x) = \sin(2x)\text{.}$ Use the double angle identity to rewrite $C$ as a product of basic functions, and use the product rule to find $C'\text{.}$ Rewrite $C'$ in the simplest form possible.

Answer

Using the double angle identity for the sine function, we write

$C(x) = \sin(2x) = 2\sin(x)\cos(x)\text{.}$

Applying the product rule and simplifying, we find

$C'(x) = 2\sin(x)(-\sin(x)) + \cos(x)(2\cos(x)) = 2(\cos^2(x) - \sin^2(x))\text{.}$

Next, we recall that the double angle identity for the cosine,

$\cos(2x) = \cos^2(x) - \sin^2(x)\text{.}$

Substituting this result into our expression for $C'(x)\text{,}$ we now have that

$C'(x) = 2 \cos(2x)\text{.}$

In Example $2.5.3$ , if we let $g(x) = 2x$ and $f(x) = \sin(x)\text{,}$ we observe that $C(x) = f(g(x))\text{.}$ Now, $g'(x) = 2$ and $f'(x) = \cos(x)\text{,}$ so we can view the structure of $C'(x)$ as

C'(x) = 2\cos(2x) = g'(x) f'(g(x))\text{.}

In this example, as in the example involving linear functions, we see that the derivative of the composite function $C(x) = f(g(x))$ is found by multiplying the derivatives of $f$ and $g\text{,}$ but with $f'$ evaluated at $g(x)\text{.}$

It makes sense intuitively that these two quantities are involved in the rate of change of a composite function: if we ask how fast $C$ is changing at a given $x$ value, it clearly matters how fast $g$ is changing at $x\text{,}$ as well as how fast $f$ is changing at the value of $g(x)\text{.}$ It turns out that this structure holds for all differentiable functions¹ as is stated in the Chain Rule.

Like other differentiation rules, the Chain Rule can be proved formally using the limit definition of the derivative.

Chain Rule

If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)\text{,}$ then the composite function $C$ defined by $C(x) = f(g(x))$ is differentiable at $x$ and

$C'(x) = f'(g(x)) g'(x)\text{.}$

As with the product and quotient rules, it is often helpful to think verbally about what the chain rule says: “If $C$ is a composite function defined by an outer function $f$ and an inner function $g\text{,}$ then $C'$ is given by the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.”

It is helpful to identify clearly the inner function $g$ and outer function $f\text{,}$ compute their derivatives individually, and then put all of the pieces together by the chain rule.

Example $2.5.4$

Determine the derivative of the function

r(x) = (\tan(x))^2\text{.}

Answer

The function $r$ is composite, with inner function $g(x) = \tan(x)$ and outer function $f(x) = x^2\text{.}$ Organizing the key information involving $f\text{,}$ $g\text{,}$ and their derivatives, we have

$f(x) = x^2$		$g(x) = \tan(x)$
$f'(x) = 2x$		$g'(x) = \sec^2(x)$
$f'(g(x)) = 2\tan(x)$

Applying the chain rule, we find that

r'(x) = f'(g(x))g'(x) = 2\tan(x) \sec^2(x)\text{.}

As a side note, we remark that $r(x)$ is usually written as $\tan^2(x)\text{.}$ This is common notation for powers of trigonometric functions: $\cos^4(x)\text{,}$ $\sin^5(x)\text{,}$ and $\sec^2(x)$ are all composite functions, with the outer function a power function and the inner function a trigonometric one.

Activity $2.5.2$

For each function given below, identify an inner function $g$ and outer function $f$ to write the function in the form $f(g(x))\text{.}$ Determine $f'(x)\text{,}$ $g'(x)\text{,}$ and $f'(g(x))\text{,}$ and then apply the chain rule to determine the derivative of the given function.

$\displaystyle h(x) = \cos(x^4)$
$\displaystyle p(x) = \sqrt{ \tan(x) }$
$\displaystyle s(x) = 2^{\sin(x)}$
$\displaystyle z(x) = \cot^5(x)$
$\displaystyle m(x) = (\sec(x) + e^x)^9$

Using multiple rules simultaneously

The chain rule now joins the sum, constant multiple, product, and quotient rules in our collection of techniques for finding the derivative of a function through understanding its algebraic structure and the basic functions that constitute it. It takes practice to get comfortable applying multiple rules to differentiate a single function, but using proper notation and taking a few extra steps will help.

Example $2.5.5$

Find a formula for the derivative of $h(t) = 3^{t^2 + 2t}\sec^4(t)\text{.}$

Answer

We first observe that $h$ is the product of two functions: $h(t) = a(t) \cdot b(t)\text{,}$ where $a(t) = 3^{t^2 + 2t}$ and $b(t) = \sec^4(t)\text{.}$ We will need to use the product rule to differentiate $h\text{.}$ And because $a$ and $b$ are composite functions, we will need the chain rule. We therefore begin by computing $a'(t)$ and $b'(t)\text{.}$

Writing $a(t) = f(g(t)) = 3^{t^2 + 2t}\text{,}$ and finding the derivatives of $f$ and $g\text{,}$ we have

$f(t) = 3^t$		$g(t) = t^2 + 2t$
$f'(t) = 3^t \ln(3)$		$g'(t) = 2t+2$
$f'(g(t)) = 3^{t^2 + 2t}\ln(3)$

Thus, by the chain rule, it follows that $a'(t) = f'(g(t))g'(t) = 3^{t^2 + 2t}\ln(3) (2t+2)\text{.}$

Turning next to $b\text{,}$ we write $b(t) = r(s(t)) = \sec^4(t)$ and find the derivatives of $r$ and $s\text{.}$

$r(t) = t^4$		$s(t) = \sec(t)$
$r'(t) = 4t^3$		$s'(t) = \sec(t)\tan(t)$
$r'(s(t)) = 4\sec^3(t)$

By the chain rule,

b'(t) = r'(s(t))s'(t) = 4\sec^3(t)\sec(t)\tan(t) = 4 \sec^4(t) \tan(t)\text{.}

Now we are finally ready to compute the derivative of the function $h\text{.}$ Recalling that $h(t) = 3^{t^2 + 2t}\sec^4(t)\text{,}$ by the product rule we have

h'(t) = 3^{t^2 + 2t} \frac{d}{dt}[\sec^4(t)] + \sec^4(t) \frac{d}{dt}[3^{t^2 + 2t}]\text{.}

From our work above with $a$ and $b\text{,}$ we know the derivatives of $3^{t^2 + 2t}$ and $\sec^4(t)\text{,}$ and therefore

h'(t) = 3^{t^2 + 2t} 4\sec^4(t) \tan(t) + \sec^4(t) 3^{t^2 + 2t}\ln(3) (2t+2)\text{.}

Activity $2.5.3$

For each of the following functions, find the function's derivative. State the rule(s) you use, label relevant derivatives appropriately, and be sure to clearly identify your overall answer.

$\displaystyle p(r) = 4\sqrt{r^6 + 2e^r}$
$\displaystyle m(v) = \sin(v^2) \cos(v^3)$
$\displaystyle h(y) = \frac{\cos(10y)}{e^{4y}+1}$
$\displaystyle s(z) = 2^{z^2 \sec (z)}$
$\displaystyle c(x) = \sin(e^{x^2})$

The chain rule now adds substantially to our ability to compute derivatives. Whether we are finding the equation of the tangent line to a curve, the instantaneous velocity of a moving particle, or the instantaneous rate of change of a certain quantity, if the function under consideration is a composition, the chain rule is indispensable.

Activity $2.5.4$

Use known derivative rules, including the chain rule, as needed to answer each of the following questions.

Find an equation for the tangent line to the curve $y= \sqrt{e^x + 3}$ at the point where $x=0\text{.}$
If $\displaystyle s(t) = \frac{1}{(t^2+1)^3}$ represents the position function of a particle moving horizontally along an axis at time $t$ (where $s$ is measured in inches and $t$ in seconds), find the particle's instantaneous velocity at $t=1\text{.}$ Is the particle moving to the left or right at that instant?
At sea level, air pressure is 30 inches of mercury. At an altitude of $h$ feet above sea level, the air pressure, $P\text{,}$ in inches of mercury, is given by the function $P = 30 e^{-0.0000323 h}\text{.}$ Compute $dP/dh$ and explain what this derivative function tells you about air pressure, including a discussion of the units on $dP/dh\text{.}$ In addition, determine how fast the air pressure is changing for a pilot of a small plane passing through an altitude of $1000$ feet.

Suppose that

f(x)

and

g(x)

are differentiable functions and that the following information about them is known:

Table $2.5.1$ . Data for functions $f$ and $g\text{.}$
$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$-1$	$2$	$-5$	$-3$	$4$
$2$	$-3$	$4$	$-1$	$2$

If $C(x)$ is a function given by the formula $f(g(x))\text{,}$ determine $C'(2)\text{.}$ In addition, if $D(x)$ is the function $f(f(x))\text{,}$ find $D'(-1)\text{.}$

The composite version of basic function rules

As we gain more experience with differentiation, we will become more comfortable in simply writing down the derivative without taking multiple steps. This is particularly simple when the inner function is linear, since the derivative of a linear function is a constant.

Example $2.5.6$

Use the chain rule to differentiate each of the following composite functions whose inside function is linear:

$\frac{d}{dx} \left[ (5x+7)^{10} \right] = 10(5x+7)^9 \cdot 5\text{,}$

$\frac{d}{dx} \left[ \tan(17x) \right] = 17\sec^2(17x), \ \text{and}$

$\frac{d}{dx} \left[ e^{-3x} \right] = -3e^{-3x}\text{.}$

More generally, following is an excellent exercise for getting comfortable with the derivative rules. Write down a list of all the basic functions whose derivatives we know, and list the derivatives. Then write a composite function with the inner function being an unknown function $u(x)$ and the outer function being a basic function. Finally, write the chain rule for the composite function. The following example illustrates this for two different functions.

Example $2.5.7$

To determine

\frac{d}{dx}[\sin(u(x))]\text{,}

where $u$ is a differentiable function of $x\text{,}$ we use the chain rule with the sine function as the outer function. Applying the chain rule, we find that

\frac{d}{dx}[\sin(u(x))] = \cos(u(x)) \cdot u'(x)\text{.}

This rule is analogous to the basic derivative rule that $\frac{d}{dx}[\sin(x)] = \cos(x)\text{.}$

Similarly, since $\frac{d}{dx}[a^x] = a^x \ln(a)\text{,}$ it follows by the chain rule that

\frac{d}{dx}[a^{u(x)}] = a^{u(x)} \ln(a) \cdot u'(x)\text{.}

This rule is analogous to the basic derivative rule that $\frac{d}{dx}[a^{x}] = a^{x} \ln(a)\text{.}$

Summary

A composite function is one where the input variable $x$ first passes through one function, and then the resulting output passes through another. For example, the function $h(x) = 2^{\sin(x)}$ is composite since $x \longrightarrow \sin(x) \longrightarrow 2^{\sin(x)}\text{.}$
Given a composite function $C(x) = f(g(x))$ where $f$ and $g$ are differentiable functions, the chain rule tells us that
$C'(x) = f'(g(x)) g'(x)\text{.}$

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2.5: The Chain Rule

Motivating Questions

Example 2.5.12.5.1

Preview Activity 2.5.12.5.1

Example 2.5.22.5.2

Example 2.5.32.5.3

Chain Rule

Example 2.5.42.5.4

Activity 2.5.22.5.2

Using multiple rules simultaneously

Example 2.5.52.5.5

Activity 2.5.32.5.3

Activity 2.5.42.5.4

The composite version of basic function rules

Example 2.5.62.5.6

Example 2.5.72.5.7

Summary

Example $2.5.1$

Preview Activity $2.5.1$

Example $2.5.2$

Example $2.5.3$

Example $2.5.4$

Activity $2.5.2$

Example $2.5.5$

Activity $2.5.3$

Activity $2.5.4$

Example $2.5.6$

Example $2.5.7$