# How do you differentiate f(x)= -1 / (x^3-4 ) using the quotient rule?

Mar 15, 2016

$f ' \left(x\right) = \frac{3 {x}^{2}}{{x}^{3} - 4} ^ 2$

#### Explanation:

The quotient rule is stated thus:

If $f \left(x\right) = \frac{u}{v}$
then $f ' \left(x\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

Let's apply that rule to our function : $f \left(x\right) = - \frac{1}{{x}^{3} - 4}$

Step 1:
$v \frac{\mathrm{du}}{\mathrm{dx}} = \left({x}^{3} - 4\right) \left(0\right) = 0$

Step 2:
$u \frac{\mathrm{dv}}{\mathrm{dx}} = \left(- 1\right) \left(3 {x}^{2}\right) = - 3 {x}^{2}$

Step 3:
${v}^{2} = {\left({x}^{3} - 4\right)}^{2}$

Step 4:
$f ' \left(x\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2 = \frac{0 - \left(- 3 {x}^{2}\right)}{{x}^{3} - 4} ^ 2 = \textcolor{b l u e}{\frac{3 {x}^{2}}{{x}^{3} - 4} ^ 2}$