How do you differentiate f(x)=e^(csc2x) using the chain rule.?

Dec 14, 2015

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

Explanation:

Using the chain rule, we get

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

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

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

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

Thus

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