# How do you differentiate f(x) = 1/(x^3-x) using the quotient rule?

Jan 11, 2016

$f ' \left(x\right) = \frac{1 - 3 {x}^{2}}{{x}^{3} - x} ^ 2$

#### Explanation:

The quotient rule states that

$\frac{d}{\mathrm{dx}} \left[\frac{g \left(x\right)}{h \left(x\right)}\right] = \frac{g ' \left(x\right) h \left(x\right) - h ' \left(x\right) g \left(x\right)}{h \left(x\right)} ^ 2$

Here, we have

$g \left(x\right) = 1$
$h \left(x\right) = {x}^{3} - x$

Differentiate the two functions:

$g ' \left(x\right) = 0$
$h ' \left(x\right) = 3 {x}^{2} - 1$

Plug these in to the quotient rule expression to see that

$f ' \left(x\right) = \frac{0 \left({x}^{3} - x\right) - \left(3 {x}^{2} - 1\right) \left(1\right)}{{x}^{3} - x} ^ 2$

Which simplifies to be

$f ' \left(x\right) = \frac{1 - 3 {x}^{2}}{{x}^{3} - x} ^ 2$