# How do you differentiate f(x)=csc(2x-x^3)  using the chain rule?

Jan 19, 2016

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

#### Explanation:

Using the chain rule :

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \csc \left(2 x - {x}^{3}\right)$

$= - \csc \left(2 x - {x}^{3}\right) \cot \left(2 x - {x}^{3}\right) \left(\frac{d}{\mathrm{dx}} 2 x - {x}^{3}\right)$

$= - \csc \left(2 x - {x}^{3}\right) \cot \left(2 x - {x}^{3}\right) \left(\left(\frac{d}{\mathrm{dx}} 2 x\right) - \left(\frac{d}{\mathrm{dx}} {x}^{3}\right)\right)$

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

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