$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 3.4: The Quotient Rule

• • Contributed by David Guichard
• Professor (Mathematics) at Whitman College

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

What is the derivative of $$(x^2+1)/(x^3-3x)$$? More generally, we'd like to have a formula to compute the derivative of $$f(x)/g(x)$$ if we already know $$f'(x)$$ and $$g'(x)$$. Instead of attacking this problem head-on, let's notice that we've already done part of the problem: $$f(x)/g(x)= f(x)\cdot(1/g(x))$$, that is, this is "really'' a product, and we can compute the derivative if we know $$f'(x)$$ and $$(1/g(x))'$$.

So really the only new bit of information we need is $$(1/g(x))'$$ in terms of $$g'(x)$$. As with the product rule, let's set this up and see how far we can get:

\eqalign{ {d\over dx}{1\over g(x)}&=\lim_{\Delta x\to0} {{1\over g(x+\Delta x)}-{1\over g(x)}\over\Delta x}\cr& =\lim_{\Delta x\to0} {{g(x)-g(x+\Delta x)\over g(x+\Delta x)g(x)}\over\Delta x}\cr& =\lim_{\Delta x\to0} {g(x)-g(x+\Delta x)\over g(x+\Delta x)g(x)\Delta x}\cr& =\lim_{\Delta x\to0} -{g(x+\Delta x)-g(x)\over \Delta x} {1\over g(x+\Delta x)g(x)}\cr& =-{g'(x)\over g(x)^2}\cr }

Now we can put this together with the product rule:

\begin{align} {d\over dx}{f(x)\over g(x)}& =f(x){-g'(x)\over g(x)^2}+f'(x){1\over g(x)} \\ &={-f(x)g'(x)+f'(x)g(x)\over g(x)^2} \\ &= {f'(x)g(x)-f(x)g'(x)\over g(x)^2}. \end{align}

Example $$\PageIndex{1}$$

Compute the derivative of $$\dfrac{x^2+1}{x^3-3x}.$$

Solution

${d\over dx}{x^2+1\over x^3-3x}={2x(x^3-3x)-(x^2+1)(3x^2-3)\over(x^3-3x)^2}= {-x^4-6x^2+3\over (x^3-3x)^2}.$

It is often possible to calculate derivatives in more than one way, as we have already seen. Since every quotient can be written as a product, it is always possible to use the product rule to compute the derivative, though it is not always simpler.

Example $$\PageIndex{2}$$

Find the derivative of $$\sqrt{625-x^2}/\sqrt{x}$$ in two ways: using the quotient rule, and using the product rule.

Solution

Quotient rule:

${d\over dx}{\sqrt{625-x^2}\over\sqrt{x}} = {\sqrt{x}(-x/\sqrt{625-x^2})-\sqrt{625-x^2}\cdot 1/(2\sqrt{x})\over x}.$

Note that we have used $$\sqrt{x}=x^{1/2}$$ to compute the derivative of $$\sqrt{x}$$ by the power rule.

Product rule:

${d\over dx}\sqrt{625-x^2} x^{-1/2} = \sqrt{625-x^2} {-1\over 2}x^{-3/2}+{-x\over \sqrt{625-x^2}}x^{-1/2}.$

With a bit of algebra, both of these simplify to

$-{x^2+625\over 2\sqrt{625-x^2}x^{3/2}}.$

Occasionally you will need to compute the derivative of a quotient with a constant numerator, like $$10/x^2$$. Of course you can use the quotient rule, but it is usually not the easiest method. If we do use it here, we get

${d\over dx}{10\over x^2}={x^2\cdot 0-10\cdot 2x\over x^4}= {-20\over x^3},$

since the derivative of 10 is 0. But it is simpler to do this:

${d\over dx}{10\over x^2}={d\over dx}10x^{-2}=-20x^{-3}.$

Admittedly, $$x^2$$ is a particularly simple denominator, but we will see that a similar calculation is usually possible. Another approach is to remember that

${d\over dx}{1\over g(x)}={-g'(x)\over g(x)^2},$

but this requires extra memorization. Using this formula,

${d\over dx}{10\over x^2}=10{-2x\over x^4}.$

Note that we first use linearity of the derivative to pull the 10 out in front.

## Contributors

• Integrated by Justin Marshall.