3.6 The chain rule

We have seen the techniques for differentiating basic functions $(x^n,\text{sin}x,\text{cos}x,\text{etc}.)$ 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 $k\left(x\right)=\sqrt{3x^2+1}.$ 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: $h\left(x\right)=\text{sin}\left(x^3\right).$ 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 $\text{sin}\left(x^3\right).$ This chain reaction gives us hints as to what is involved in computing the derivative of $\text{sin}\left(x^3\right).$ 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 $h\left(x\right)=\text{sin}\left(x^3\right)$ by setting up the limit that would give us the derivative at a specific value $a$ in the domain of $h\left(x\right)=\text{sin}\left(x^3\right).$

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

In other words, if $h\left(x\right)=\text{sin}\left(x^3\right),$ then $h^{\prime }\left(x\right)=\text{cos}\left(x^3\right)·3x^2.$ 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 $f\left(x\right)=\text{sin}x$ evaluated at the function $g\left(x\right)=x^3.$ At this point, we anticipate that for $h\left(x\right)=\text{sin}\left(g\left(x\right)\right),$ it is quite likely that $h^{\prime }(x)=\text{cos}(g(x))g^{\prime }(x).$ As we determined above, this is the case for $h\left(x\right)=\text{sin}\left(x^3\right).$