$\sqrt{\mathrm{Tany}}={\mathrm{e}}^{\cos 2\mathrm{x}} \mathrm{sinx}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}=$

The given equation $\sqrt{\tan y}=e^{\cos 2x}$

Apply logarithm on both sides

$\begin{array}{rcl}\frac{1}{2}\log \left(\tan y\right)& =& \cos 2x+\log \sin x\\ \log \left(\tan y\right)& =& 2\cos 2x+2\log \sin x\end{array}$

Differentiate both sides

$\begin{array}{rcl}\frac{1}{\tan y}se{c}^{2}y\left(\frac{dy}{dx}\right)& =& 4\left(-\sin 2x\right)+2cotx\\ \frac{dy}{dx}& =& \left(2 cotx-4\sin 2x\right)\left(\frac{\tan y}{se{c}^{2}y}\right)\\ & =& \left(cotx-2\sin 2x\right)\left(2\sin y\cos y\right)\\ & =& \left(cotx-2\sin 2x\right)\sin 2y\end{array}$