# 3.6 The chain rule

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• State the chain rule for the composition of two functions.
• Apply the chain rule together with the power rule.
• Apply the chain rule and the product/quotient rules correctly in combination when both are necessary.
• Recognize the chain rule for a composition of three or more functions.
• Describe the proof of the chain rule.

We have seen the techniques for differentiating basic functions $\left({x}^{n},\text{sin}\phantom{\rule{0.1em}{0ex}}x,\text{cos}\phantom{\rule{0.1em}{0ex}}x,\text{etc}.\right)$ as well as sums, differences, products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate compositions of functions, such as $h\left(x\right)=\text{sin}\left({x}^{3}\right)$ or $k\left(x\right)=\sqrt{3{x}^{2}+1}.$ In this section, we study the rule for finding the derivative of the composition of two or more functions.

## Deriving the chain rule

When we have a function that is a composition of two or more functions, we could use all of the techniques we have already learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are able to differentiate can get cumbersome. Instead, we use the chain rule    , which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

To put this rule into context, let’s take a look at an example: $h\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right).$ We can think of the derivative of this function with respect to x as the rate of change of $\text{sin}\left({x}^{3}\right)$ relative to the change in $x.$ Consequently, we want to know how $\text{sin}\left({x}^{3}\right)$ changes as $x$ changes. We can think of this event as a chain reaction: As $x$ changes, ${x}^{3}$ changes, which leads to a change in $\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right).$ This chain reaction gives us hints as to what is involved in computing the derivative of $\phantom{\rule{0.2em}{0ex}}\text{sin}\left({x}^{3}\right).$ First of all, a change in $x$ forcing a change in ${x}^{3}$ suggests that somehow the derivative of ${x}^{3}$ is involved. In addition, the change in ${x}^{3}$ forcing a change in $\text{sin}\left({x}^{3}\right)$ suggests that the derivative of $\text{sin}\left(u\right)$ with respect to $u,$ where $u={x}^{3},$ is also part of the final derivative.

We can take a more formal look at the derivative of $h\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right)$ by setting up the limit that would give us the derivative at a specific value $a$ in the domain of $h\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right).$

${h}^{\prime }\left(a\right)=\underset{x\to a}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right)-\text{sin}\left({a}^{3}\right)}{x-a}.$

This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression ${x}^{3}-{a}^{3}$ to obtain

${h}^{\prime }\left(a\right)=\underset{x\to a}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right)-\text{sin}\left({a}^{3}\right)}{{x}^{3}-{a}^{3}}·\frac{{x}^{3}-{a}^{3}}{x-a}.$

From the definition of the derivative, we can see that the second factor is the derivative of ${x}^{3}$ at $x=a.$ That is,

$\underset{x\to a}{\text{lim}}\frac{{x}^{3}-{a}^{3}}{x-a}=\frac{d}{dx}\left({x}^{3}\right)=3{a}^{2}.$

However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting $u={x}^{3}$ and observing that as $x\to a,u\to {a}^{3}\text{:}$

$\begin{array}{cc}\hfill \underset{x\to a}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right)-\text{sin}\phantom{\rule{0.1em}{0ex}}\left({a}^{3}\right)}{{x}^{3}-{a}^{3}}& =\underset{u\to {a}^{3}}{\text{lim}}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}u-\text{sin}\phantom{\rule{0.1em}{0ex}}\left({a}^{3}\right)}{u-{a}^{3}}\hfill \\ & =\frac{d}{du}{\left(\text{sin}\phantom{\rule{0.1em}{0ex}}u\right)}_{u={a}^{3}}\hfill \\ & =\text{cos}\phantom{\rule{0.1em}{0ex}}\left({a}^{3}\right).\hfill \end{array}$

Thus, ${h}^{\prime }\left(a\right)=\text{cos}\left({a}^{3}\right)·3{a}^{2}.$

In other words, if $h\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right),$ then ${h}^{\prime }\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right)·3{x}^{2}.$ Thus, if we think of $h\left(x\right)=\text{sin}\left({x}^{3}\right)$ as the composition $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ where $f\left(x\right)=$ sin $x$ and $g\left(x\right)={x}^{3},$ then the derivative of $h\left(x\right)=\text{sin}\left({x}^{3}\right)$ is the product of the derivative of $g\left(x\right)={x}^{3}$ and the derivative of the function $f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ evaluated at the function $g\left(x\right)={x}^{3}.$ At this point, we anticipate that for $h\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right),$ it is quite likely that ${h}^{\prime }\left(x\right)=\text{cos}\left(g\left(x\right)\right){g}^{\prime }\left(x\right).$ As we determined above, this is the case for $h\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right).$

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