# How do you differentiate f(x)=sqrtcsc(e^(4x)) using the chain rule.?

Mar 12, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cot \left({e}^{4 x}\right) \sqrt{\csc \left({e}^{4 x}\right)}}{2}$ $4 {e}^{4 x}$

#### Explanation:

let $u = {e}^{4 x}$ so that $\frac{\mathrm{du}}{\mathrm{dx}} = 4 {e}^{4 x}$

and let $y = f \left(x\right)$ so that $y = {\left(\csc \left(u\right)\right)}^{\frac{1}{2}}$ so$\left(\frac{\mathrm{dy}}{\mathrm{du}}\right) = - \frac{\cot \left(u\right) \sqrt{\csc \left(u\right)}}{2}$

then by the chain rule

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

so
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cot \left({e}^{4 x}\right) \sqrt{\csc \left({e}^{4 x}\right)}}{2}$ $4 {e}^{4 x}$