If $\sqrt{\tan y}=e^{\cos 2x}\sin x$, then $\frac{dy}{dx}=$

The correct option is B $\sin 2y(\cot x-2\sin 2x)$
$\sqrt{tany}=e^{cos2x}\sin x$

differentiating w.r.t. $x$

$\frac{1}{2\sqrt{tany}}\times {\sec }^{2}y\frac{dy}{dx}=e^{cos2x}\times -2sin2x.\sin x+e^{cos2x}\cos x$

$\frac{{\sec }^{2}y}{2\sqrt{tany}}\frac{dy}{dx}=e^{cos2x}\left[-2sin2x.\sin x+\cos x\right]$

$\frac{{\sec }^{2}y}{2\sqrt{tany}}\frac{dy}{dx}=\frac{\sqrt{tany}}{\sin x}\left[\cos x-2sin2x.\sin x\right]$

$\frac{dy}{dx}=\frac{2\tan y{\cos }^{2}y}{\sin x}\left[\cos x-2\sin x.sin2x\right]$

$\frac{dy}{dx}=\frac{2\sin y\cos y}{\sin x}\left[\cos x-2\sin x.sin2x\right]$

$\frac{dy}{dx}=\sin 2y\left[\cot x-2sin2x\right]$