# How do you differentiate f(x)=cos(e^(3x^3-x))  using the chain rule?

Jun 21, 2016

$f ' \left(x\right) = - \left(9 {x}^{2} - 1\right) {e}^{3 {x}^{3} - x} \sin \left({e}^{3 {x}^{3} - x}\right)$

#### Explanation:

Using the sum rule and the chain rule, along with the following derivatives:

• $\frac{d}{\mathrm{dx}} \cos \left(x\right) = - \sin \left(x\right)$

• $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

• $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

we have

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \cos \left({e}^{3 {x}^{3} - x}\right)$

$= - \sin \left({e}^{3 {x}^{3} - x}\right) \left(\frac{d}{\mathrm{dx}} {e}^{3 {x}^{3} - x}\right)$

$= - \sin \left({e}^{3 {x}^{3} - x}\right) {e}^{3 {x}^{3} - x} \left(\frac{d}{\mathrm{dx}} \left(3 {x}^{3} - x\right)\right)$

$= - \sin \left({e}^{3 {x}^{3} - x}\right) {e}^{3 {x}^{3} - x} \left(\left(\frac{d}{\mathrm{dx}} 3 {x}^{3}\right) - \left(\frac{d}{\mathrm{dx}} x\right)\right)$

$= - \sin \left({e}^{3 {x}^{3} - x}\right) {e}^{3 {x}^{3} - x} \left(9 {x}^{2} - 1\right)$

$\therefore f ' \left(x\right) = - \left(9 {x}^{2} - 1\right) {e}^{3 {x}^{3} - x} \sin \left({e}^{3 {x}^{3} - x}\right)$