# Question #2ef3f

Oct 10, 2016

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

#### Explanation:

Use the chain rule to differentiate ${\sin}^{3} x = {\left(\sin x\right)}^{3}$

$f ' \left(x\right) = 3 {\sin}^{2} x \cos x - 3 \cos x$ which can be rewritten

$= 3 \cos x \left({\sin}^{2} x - 1\right)$

$= - 3 \cos x \left(1 - {\sin}^{2} x\right)$

$= - 3 \cos x \left({\cos}^{2} x\right)$

$= - 3 {\cos}^{3} x$