5.4 Product Rule and Quotient Rule

Now that we know some basic rules of derivatives, let’s move on to finding the derivative of a function consisting of multiple functions. Suppose that we have a function $y=f(x)\cdot g(x)$ where $f(x)$ and $g(x)$ are functions of $x$. How can we find the derivative? Applying the definition of a derivative,

$\begin{array}{rl}{y}^{\prime }& =\underset{h\to 0}{lim}\frac{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{f(x+h)\cdot g(x+h)-f(x)\cdot g(x+h)+f(x)\cdot g(x+h)-f(x)\cdot g(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{g(x+h)[f(x+h)-f(x)]}{h}+\underset{h\to 0}{lim}\frac{f(x)[g(x+h)-g(x)]}{h}\\ & =\underset{h\to 0}{lim}g(x+h)\frac{f(x+h)-f(x)}{h}+\underset{h\to 0}{lim}f(x)\frac{g(x+h)-g(x)}{h}\\ & =g(x)\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}+f(x)\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}\\ & =g(x)f^{\prime }(x)+f(x)g^{\prime }(x)\end{array}$

Here is the definition of the product rule again. Suppose $f(x)$ and $g(x)$ are differentiable functions, and suppose $y=f(x)$. Then the derivative of the function $y$ with respect to $x$ is

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

In words, if a function is a product of two functions of $x$, then the derivative with respect to $x$ is the derivative of the first function times the second, plus the first function times the derivative of the second function. Let’s go through some examples. Suppose we want to find the derivative of the function

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

Let $g(x)=x^2+3x-4$ and $h(x)=x^3-3x-1$. Using the power rule, the derivative of each function is as follows:

$g^{\prime }(x)=2x+3$

$h^{\prime }(x)=3x^2-3$

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

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

We have not simplified the resulting functions here, but you could simplify the functions to make them look nicer.

Now, let’s find the derivative of the function

$f(x)=e^x(x^2+3x-5)$

Let $g(x)=e^x$ and $h(x)=x^2+3x-5$. Using the power rule, the derivative of each function is as follows:

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

$h^{\prime }(x)=2x+3$

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

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

Now, let’s think about a function that represents the division of two functions such that

$y=\frac{f(x)}{g(x)}$

We can apply the definition of a derivative to get

$\begin{array}{rl}{y}^{\prime }& =\underset{h\to 0}{lim}\frac{f(x+h)/g(x+h)-f(x)/g(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{f(x+h)g(x)/g(x+h)g(x)-f(x)g(x+h)/g(x+h)g(x)}{h}\\ & =\underset{h\to 0}{lim}\left(\frac{1}{g(x)g(x+h)}\right)\left(\frac{f(x+h)g(x)-f(x)g(x+h)}{h}\right)\\ & =\frac{1}{[g(x){]}^2}\underset{h\to 0}{lim}\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{h}\\ & =\frac{1}{[g(x){]}^2}\underset{h\to 0}{lim}g(x)\frac{f(x+h)-f(x)}{h}-\underset{h\to 0}{lim}f(x)\frac{g(x+h)-g(x)}{h}\\ & =\frac{1}{[g(x){]}^2}(f^{\prime }(x)g(x)-f(x)g^{\prime }(x))\\ & =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}\end{array}$

Given this result, suppose $f(x)$ and $g(x)$ are differentiable functions. Define $y=\frac{f(x)}{g(x)}$  and $g(x)\ne 0$. Then

$y^{\prime }=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}$

In words, if a function is a division of one function over another, then the derivative of the function is the derivative of the numerator function times the denominator function, minus the numerator function times the derivative of the denominator function, all divided by the square of the denominator function. Let’s practice the quotient rule. Suppose we want to find the derivative of the following function:

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

Let $g(x)=x^2-1$ and $h(x)=x+2$. Using the power rule, the derivative of each function is as follows:

$g^{\prime }(x)=2x$

$h^{\prime }(x)=1$

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

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

Even though we have not simplified here, you could always simplify and clean up the function.

How about the derivative of the following function?

$f(x)=\frac{x^3}{e^x}$

Let $g(x)=x^3$ and $h(x)=e^x$. Using the power rule, the derivative of each function is as follows:

$g^{\prime }(x)=3x^2$

$h^{\prime }(x)=e^x$

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

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{g^{\prime }(x)h(x)-g(x)h^{\prime }(x)}{[h(x){]}^2}\\ & =\frac{(3x^2)e^x-x^3{e}^x}{(e^x{)}^2}\\ & =\frac{(3x^2-x^3)e^x}{e^{2x}}\end{array}$