# How do you differentiate f(x)=e^(csc(2/x) using the chain rule.?

Jul 13, 2016

This will require multiple applications of the chain rule.

Let's start with $\csc \left(\frac{2}{x}\right)$. Let $y = \csc u$ and $u = \frac{2}{x}$.

Let's differentiate:

$y = \frac{1}{\sin} u$

$y ' = \frac{\left(0 \times \sin u\right) - \left(1 \times \cos u\right)}{\sin u} ^ 2$

$y ' = - \cos \frac{u}{\sin} ^ 2 u$

$y ' = - \cot u \csc u$

$u ' = \frac{\left(0 \times x\right) - \left(2 \times 1\right)}{x} ^ 2$

$u ' = - \frac{2}{x} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{x} ^ 2 \times - \cot u \csc u$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{- 2 \cot u \csc u}{x} ^ 2$

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

Now that we know this derivative we can say:

let $y = {e}^{u}$ and $u = \frac{2 \cot \left(\frac{2}{x}\right) \csc \left(\frac{2}{x}\right)}{x} ^ 2$

We already know the derivative of $u$ and the derivative of $y$ is ${e}^{u}$.

Hence:

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

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

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

Hopefully this helps!