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OpenStudy (anonymous):
xsin2y=ycos2x help me to differentiate this plzz
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OpenStudy (anonymous):
siny+xcos2y*2y'=y'cos2x-y2sin2x
OpenStudy (lalaly):
so \[2y'xcos(2y)-y'\cos(2x)=-2ysin(2x)-\sin(2y)\]\[y'(2xcos(2y)-\cos(2x))=-2ysin(2x)-\sin(2y)\]\[y'=\frac{-2ysin(2x)-\sin(2y)}{2xcos(2y)-\cos(2x)}\]
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