# Question #6bed4

##### 1 Answer
Jan 23, 2017

$\left(- 6 {t}^{2} \sin \left({t}^{3}\right)\right) ' = - 12 t \sin \left({t}^{3}\right) - 18 {t}^{4} \cos \left({t}^{3}\right)$

#### Explanation:

First remember the product rule

$\left(f \left(x\right) g \left(x\right)\right) ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

In this case

$f \left(x\right) = - 6 {t}^{2}$

and

$g \left(x\right) = \sin \left({t}^{3}\right)$

then

since

$\left(c {x}^{n}\right) ' = c \left({x}^{n}\right) ' = c n {x}^{n - 1}$ power rule and derivative times a constant equals constant times derivative

$f ' \left(x\right) = - 12 t$

also by the chain rule $\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

$g ' \left(x\right) = 3 {t}^{2} \cos \left({t}^{3}\right)$

Then

$\left(- 6 {t}^{2} \sin \left({t}^{3}\right)\right) ' = - 12 t \sin \left({t}^{3}\right) - 18 {t}^{4} \cos \left({t}^{3}\right)$