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

Mar 1, 2017

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

#### Explanation:

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{2}{2}} |}}} \leftarrow \text{ chain rule}$

$\text{let } u = {e}^{x} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = {e}^{x}$

$\Rightarrow y = \csc u \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = - \csc u \cot u$

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

change u back into terms of x

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