# How do you differentiate f(x)=csc(2x -4)  using the chain rule?

Jan 8, 2016

$f ' \left(x\right) = - 2 \csc \left(2 x - 4\right) \cot \left(2 x - 4\right)$

#### Explanation:

According to the chain rule,

$\frac{d}{\mathrm{dx}} \left(\csc \left(u\right)\right) = - \csc \left(u\right) \cot \left(u\right) \cdot u '$

Thus we have

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\csc \left(2 x - 4\right)\right) = - \csc \left(2 x - 4\right) \cot \left(2 x - 4\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x - 4\right)$

$= - 2 \csc \left(2 x - 4\right) \cot \left(2 x - 4\right)$