3.4: The Quotient Rule

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:

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

Now we can put this together with the product rule:

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

Example $\PageIndex{1}$

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

Solution

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

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:

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

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

Product rule:

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

With a bit of algebra, both of these simplify to

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

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

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

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

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

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

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

but this requires extra memorization. Using this formula,

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

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

Contributors

• David Guichard (Whitman College)

• Integrated by Justin Marshall.