Chain rule

Section 15 Chain rule

Objectives

Understand and apply the chain rule for computing derivatives of compositions of functions.
Incorporate the chain rule with our existing rules and formulas to compute derivatives of complicated functions.

Our current list of derivative rules take into account how the derivative operation interacts with various function arithmetic operations: e.g., addition, scalar multiplication, product, quotient. We round out this discussion by considering another very important function operation: namely, function composition.

Theorem 15.1. Chain rule.

Assume that the function $f$ is defined as the composition of the functions $g$ and $u\text{,}$ so that $f(x)=g(u(x))\text{.}$ If $u$ is differentiable at $x\text{,}$ and $g$ is differentiable at $u(x)\text{,}$ then $f$ is differentiable at $x\text{,}$ and we have

\begin{align} f'(x) \amp = g'(u(x))u'(x)\text{.}\tag{15.1} \end{align}

Alternatively, using Leibniz notation, we have

\begin{align} \frac{df}{dx} \amp = \frac{df}{du}\frac{du}{dx}\text{.}\tag{15.2} \end{align}

The main difficulty in using the chain rule is being able to recognize how a given function $f$ can be described as a composition of functions $f(x)=g(u(x))\text{.}$ The procedure below is designed to help you identify the inner function $u$ in this representation. You do not need to use it when computing derivatives with the chain rule, but it may help you get the hang of this technique. See the first computation in Example 15.3 for an illustration of Procedure 15.2.

Procedure 15.2. Chain rule.

To compute the derivative of a function $f$ that is built from other functions using composition, proceed as follows.

Identify inner function.
Identify a function $u=u(x)$ so that $f(x)=g(u(x))$ for some function $g\text{.}$ The function $u(x)$ might be easily identified as an “inner function”, or may be a common expression that appears in the definition of $f\text{.}$
Assemble ingredients.

Compute

\begin{align*} \frac{dg}{du} \amp = g'(u)\\ \frac{du}{dx} \amp = u'(x)\text{,} \end{align*}

where $g'(u)$ is computed treating $u$ simply as a new variable name.
Use chain rule.

Compute

\begin{align*} f'(x) \amp = g'(u) u'(x)\\ \amp = g'(u(x))u'(x)\text{,} \end{align*}

making sure to leave your final answer entirely in terms of $x\text{.}$

Example 15.3. Chain rule.

Compute the derivative of the given function. Your answer should be a chain of equalities with steps justified.

$\displaystyle h(s)=\sin(2s)$
$\displaystyle f(x)=\frac{1}{x^2+2x+3}$
$\displaystyle g(x)=\sec(\sqrt{x})$
$\displaystyle u(t)=(t^3-t+1)^{100}\sqrt{\tan t}$

Solution.

We will illustrate Procedure 15.2 for this example. We declare

\begin{equation*} u=2s\text{,} \end{equation*}

from whence we have

\begin{align*} f(s) \amp = \sin u\\ u'(s) \amp = 2\text{.} \end{align*}

We now compute

\begin{align*} f'(s) \amp = \frac{d}{du}(\sin u)\cdot u'(s) \amp \text{(chain rule)}\\ \amp = \cos u\cdot 2\\ \amp = 2\cos(2s)\text{.} \end{align*}

Note that we are careful to leave our final answer entirely in terms of the original variable.
That this is a chain rule computation is more clearly seen after a little algebra:

\begin{equation*} f(x)=\frac{1}{x^2+2x+3}=(x^2+2x+3)^{-1}\text{.} \end{equation*}

We now compute

\begin{align*} f'(x) \amp = (-1)(x^2+2x+3)^{-2}\cdot (x^2+2x+3)' \amp \text{(chain rule)}\\ \amp = -(x^2+2x+3)^{-2}(2x+2)\\ \amp = -\frac{2x+2}{(x^2+2x+3)^2}\text{.} \end{align*}
We compute:

\begin{align*} u'(t) \amp = \frac{d}{dt}[(t^3-t+1)^{100}]\cdot \sqrt{\tan t}+(t^3-t+1)^{100}\cdot \frac{d}{dt}[(\tan t)^{1/2}] \amp \text{(prod. rule)}\\ \amp = 100(t^3-t+1)^{99}(t^3-t+1)'\cdot \sqrt{\tan t}+(t^3-t+1)^{100}\cdot \frac{1}{2}(\tan t)^{-1/2}(\tan t)' \amp \text{(chain rule)}\\ \amp = 100(t^3-t+1)^{99}(3t^2-1)\cdot \sqrt{\tan t}+\frac{1}{2}(t^3-t+1)^{100}(\tan t)^{-1/2}\sec^2 t\\ \amp = (t^3-t+1)^{99}\left(100(3t^2-1)\sqrt{\tan t}+\frac{(t^3-t+1)\sec^2 t}{2\sqrt{\tan t}}\right)\text{.} \end{align*}

Example 15.4. Inflating a balloon.

The volume $V$ (in cm$^3$) of a spherical inflatable balloon is computed as $V=\frac{4}{3}\pi r^3\text{,}$ where $r$ is the radius of the balloon (in cm).

Compute the rate of change of the volume $V$ with respect to the radius $r\text{.}$
Suppose now that while inflating the balloon, the radius $r$ is given by the function $r=h(t)\text{.}$ Compute the rate of change of $V$ with respect to $t\text{.}$ Leave your answer in terms of $h$ and $h'\text{.}$

Solution.

We have

\begin{align*} \frac{dV}{dr} \amp =\frac{4}{3}\pi \frac{d}{dr}(r^3) \amp \text{(scal. mult.)}\\ \amp = 4\pi\, r^2 \text{.} \end{align*}
When $r=h(t)$ is a function of time, the volume $V$ of the ballon is also a function of time:

\begin{equation*} V=V(t)=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (h(t))^3\text{.} \end{equation*}

Using the chain rule, we have

\begin{align*} \frac{dV}{dt} \amp = \frac{4}{3}\pi \frac{d}{dt}((h(t)^3)' \amp \text{(scal. mult.)}\\ \amp =\frac{4}{3}\pi \cdot 3 (h(t))^2\, h'(t) \amp \text{(chain)}\\ \amp =\underset{\frac{dV}{dr}}{\underbrace{4\pi r^2}}\cdot \underset{\frac{dr}{dt}}{\underbrace{h'(t)}} \text{.} \end{align*}

We thus see that at any given time $t_0\text{,}$ the rate of change with respect to $t$ is the product of the rate of change of volume with respect to $r$ for the radius value at $t_0$ (i.e., for $r_0=h(t_0)$) and the rate of change of $r$ with respect to time at that time $t_0\text{.}$ (Of course this is just as the chain rule predicts: $\frac{dV}{dt}=\frac{dV}{dr}\cdot \frac{dr}{dt}$.) This makes general sense, physically. The rate of change of the radius is acting as a multiplier for the rate of change of the volume with respect to $t\text{:}$ in particular the greater the rate at which the radius increases, the greater the rate at which the volume increases.

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