# How do you differentiate y = 6 cos(x^3 + 3)?

##### 1 Answer
Mar 14, 2017

$y ' = - 18 {x}^{2} \sin \left({x}^{3} + 3\right)$

#### Explanation:

Use $\left(\cos u\right) ' = - \sin u \left(u '\right)$

Let $u = {x}^{3} + 3$, $u ' = 3 {x}^{2}$

If $y = 6 \cos \left({x}^{3} + 3\right)$ then

$y ' = 6 \left(- \sin \left({x}^{3} + 3\right)\right) \left(3 {x}^{2}\right)$

Simplify: $y ' = - 18 {x}^{2} \sin \left({x}^{3} + 3\right)$