# How do you differentiate f(x)=1/(x^3-1)*x^2*cosx using the product rule?

Jul 4, 2018

$f ' \left(x\right) = - \frac{{x}^{4} + 2 x}{{x}^{3} - 1} ^ 2 \cdot \cos \left(x\right) - {x}^{2} / \left({x}^{3} - 1\right) \cdot \sin \left(x\right)$

#### Explanation:

Using the product rule

$\left(u v\right) ' = u ' v + u v '$ we get with
$u = {x}^{2} / \left({x}^{3} - 1\right)$

$u ' = \frac{2 x \left({x}^{3} - 1\right) - {x}^{2} \cdot 3 {x}^{2}}{{x}^{3} - 1} ^ 2$
simplifying we get

$u ' = - \frac{{x}^{4} + 2 x}{{x}^{3} - 1} ^ 2$
$v = \cos \left(x\right)$
$v ' = - \sin \left(x\right)$

putting thins together we get

$f ' \left(x\right) = - \frac{{x}^{4} + 2 x}{{x}^{3} - 1} ^ 2 \cos \left(x\right) - {x}^{2} / \left({x}^{3} - 1\right) \sin \left(x\right)$