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We have seen the techniques for differentiating basic functions $({x}^{n},\text{sin}\phantom{\rule{0.1em}{0ex}}x,\text{cos}\phantom{\rule{0.1em}{0ex}}x,\text{etc}.)$ 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.
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 $\text{sin}(u)$ 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 $h\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right).$
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
From the definition of the derivative, we can see that the second factor is the derivative of ${x}^{3}$ at $x=a.$ That is,
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 $x\to a,u\to {a}^{3}\text{:}$
Thus, ${h}^{\prime}\left(a\right)=\text{cos}\left({a}^{3}\right)\xb73{a}^{2}.$
In other words, if $h\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right),$ then ${h}^{\prime}\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}\right)\xb73{x}^{2}.$ 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 ${h}^{\prime}(x)=\text{cos}(g(x)){g}^{\prime}(x).$ 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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