# How do you implicitly differentiate y sin 2x = x cos 2y?

It is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos 2 y - 2 y \cos 2 x}{\sin 2 x + 2 x \sin 2 y}$
dy/dx*sin2x+2*y*cos2x=cos2y-2xsin2y*dy/dx=> dy/dx*[sin2x+2xsin2y]=cos2y-2ycos2x=> dy/dx=[cos2y-2ycos2x]/[sin2x+2xsin2y]