Question

xsin2y=ycos2x help me to differentiate this plzz

Answer 1

siny+xcos2y*2y'=y'cos2x-y2sin2x

Answer 2

so \[2y'xcos(2y)-y'\cos(2x)=-2ysin(2x)-\sin(2y)\]\[y'(2xcos(2y)-\cos(2x))=-2ysin(2x)-\sin(2y)\]\[y'=\frac{-2ysin(2x)-\sin(2y)}{2xcos(2y)-\cos(2x)}\]