5.5 Chain Rule

Suppose we want to find the derivatives of the following functions:

$f(x)=\sqrt{x^2-1}$

$f(x)=2^{x^3+x}$

$f(x)=e^{2x^3-3x^2-1}$

$f(x)=\ln (3x^2-3)$

How can we find the derivative of these functions with respect to $x$? Unfortunately, none of the rules we learned in this topic will help us find the derivatives. These functions can be written as $g(h(x))$. For example, with the first one, we can say $g(x)=\sqrt{x}$ and replace the $x$ of the $g$ function with $h(x)=x^2-1$. Can you identify the two functions that compose each of the other functions?

$g(x)=\sqrt{x},h(x)=x^2-1$
$g(x)=2^x,h(x)=x^3+x$
$g(x)=e^x,h(x)=2x^3-3x^2-1$
$g(x)=\ln x,h(x)=3x^2-3$

This illustrates the last rule of derivatives—the chain rule. If $h(x)$ is differentiable at $x$ and $g(x)$ is differentiable at $h(x)$, then $f(x)=g(h(x))$ is differentiable at $x$, and the derivative of $f(x)$ can be found by

$f^{\prime }(x)=h^{\prime }(x)\cdot g^{\prime }(h(x))$

Let’s use the original four equations to find the derivative of each of the others as practice. First,

$f(x)=\sqrt{x^2-1}$

can be decomposed as follows:

$\begin{array}{rl}g(x)& =\sqrt{x},\\ h(x)& =x^2-1\end{array}$

Let’s find the derivative of the functions $g$ and $h$ first.

$\begin{array}{rl}{g}^{\prime }(x)& =\frac{1}{2\sqrt{x}},\\ h^{\prime }(x)& =2x\end{array}$

Therefore, the derivative of the function $f(x)$ is

$\begin{array}{rl}{f}^{\prime }(x)& =h^{\prime }(x)\cdot g^{\prime }(h(x))\\ & =2x\cdot \frac{1}{2\sqrt{x^2-1}}\\ & =\frac{x}{\sqrt{x^2-1}}\end{array}$

Given the above result, the general formula to find the derivative of a function $f(x)$ that is some constant $n$th power of a function $g(x)$, expressed as $f(x)=[g(x){]}^n$, is

$f^{\prime }(x)=n{g}^{\prime }(x)[g(x){]}^{n-1}$

Suppose we want to find the derivative of

$f(x)=(x^3-2x^2-x{)}^3$

The derivative of the function is

$\begin{array}{rl}{f}^{\prime }(x)& =n{g}^{\prime }(x)[g(x){]}^{n-1}\\ & =3(3x^2-4x-1)(x^3-2x^2-x{)}^2\end{array}$

Next, instead of looking at the second function, let’s skip to the third function.

$f(x)=e^{2x^3-3x^2-1}$

This function is composed of the following two functions:

$\begin{array}{rl}g(x)& =e^x\\ h(x)& =2x^3-3x^2-1\end{array}$

The derivatives of these two functions are

$\begin{array}{rl}{g}^{\prime }(x)& =e^x\\ h^{\prime }(x)& =6x^2-6x\end{array}$

Therefore, the derivative of $f$ with the chain rule is

$\begin{array}{rl}{f}^{\prime }(x)& =h^{\prime }(x)\cdot g^{\prime }(h(x))\\ & =(6x^2-6x)e^{2x^3-3x^2-1}\end{array}$

Given this result, we can get a general formula of the derivative of a function $f$ that is the constant $e$ to the $g(x)$ power as follows:

$f^{\prime }(x)=g^{\prime }(x)e^{g(x)}$

Let’s look at the second example now. For this, we have to remember many rules and equations.

$f(x)=2^{x^3+x}$

As we found earlier, the composition of the function $f$ is

$g(x)=2^x$
$h(x)=x^3+x$

Before we tackle this, we need to rewrite $g(x)$ as follows:

$\begin{array}{rl}g(x)& =2^x\\ & =e^{x\ln 2}(\text{Recall}{e}^{\ln a}=a)\end{array}$

So the derivative of $g$ needs the chain rule as well. Let

$\begin{array}{rl}m(x)& =e^x\\ n(x)& =x\ln 2\end{array}$

The derivatives of the functions $m$ and $n$ are

$\begin{array}{rl}{m}^{\prime }(x)& =e^x\\ n^{\prime }(x)& =\ln 2\end{array}$

So the derivatives of the function $g$ and $h$ are

$\begin{array}{rl}{g}^{\prime }(x)& =n^{\prime }(x)\cdot m^{\prime }(n(x))\\ & =\ln 2\cdot e^{x\ln 2}\\ & =2^x\ln 2\\ h^{\prime }(x)& =3x^2+1\end{array}$

Therefore, the derivative of $f$ is

$\begin{array}{rl}{f}^{\prime }(x)& =h^{\prime }(x)\cdot g^{\prime }(h(x))\\ & =(3x^2+1)2^{x^3+x}\ln 2\end{array}$

So, the general formula of the derivative of the function $f$ such that constant $a$ to the power of a function $g(x)(f(x=a^{g(x)})$ is

$f^{\prime }(x)=g^{\prime }(x)a^{g(x)}\ln a$

Here is a practice problem. What is the derivative of $f(x)={10}^{x^2-3}$? First, we can identify $g(x)$ and $g^{\prime }(x)$.

$\begin{array}{rl}g(x)& =x^2-3\\ g^{\prime }(x)& =2x\end{array}$

Therefore, the derivative of $f$ is

$\begin{array}{rl}{f}^{\prime }(x)& =g^{\prime }(x)a^{g(x)}\ln a\\ & =2x\cdot {10}^{x^2-3}\ln 10\end{array}$

Here is the last one:

$f(x)=\ln (3x^2-3)$

As we found earlier, the components of $f$ are

$\begin{array}{rl}g(x)& =\ln x\\ h(x)& =3x^2-3\end{array}$

The derivatives of the functions $g$ and $h$ are

$\begin{array}{rl}{g}^{\prime }(x)& =\frac{1}{x}\\ h^{\prime }(x)& =6x\end{array}$

Therefore, the derivative is

$\begin{array}{rl}{f}^{\prime }(x)& =h^{\prime }(x)\cdot g^{\prime }(h(x))\\ & =6x\cdot \frac{1}{3x^2-3}\\ & =\frac{2x}{x^2-1}\end{array}$

Let’s practice. Suppose that we want to find the derivative of the following function:

$f(x)=\ln (x^4-3x^2+2x-1)$

Since

$\begin{array}{rl}g(x)& =x^4-3x^2+2x-1\\ g^{\prime }(x)& =4x^3-6x+2\end{array}$

then the derivative of $f$ is

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{g^{\prime }(x)}{g(x)}\\ & =\frac{4x^3-6x+2}{x^4-3x^2+2x-1}\end{array}$