If $\tan \mathrm{y}={\mathrm{e}}^{\cos 4\mathrm{x}}\cdot \sin 2\mathrm{x}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}=$

The given equation is $\tan y=e^{\cos 4x}\sin 2x$

Differentiate both sides 

$\begin{array}{rcl}se{c}^{2}y\left(\frac{dy}{dx}\right)& =& e^{\cos 4x}\left(2\cos 2x\right)+\sin \left(2x\right)\left(e^{\cos 4x}\right)\left(-4\sin 4x\right)\\ & =& 2e^{\cos 4x}\sin 2x\left(cot2x-2\sin 4x\right)\\ \frac{dy}{dx}& =& 2\tan yse{c}^{2}y\left(cot2x-2\sin 4x\right)\\ & =& \sin 2y\left(cot2x-2\sin 4x\right)\end{array}$