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  • Find dy/dx  in the following: sin ^2 y+ cos xy = k

7. Find dy/dx  in the following: \sin ^ 2 y + \cos xy = k

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Given function is
\sin ^ 2 y + \cos xy = k
Now, differentiation w.r.t. x is
\frac{d(\sin^2y+\cos xy)}{dx} = \frac{d(k)}{dx}
2\sin y \cos y\frac{dy}{dx}+(-\sin xy)(y+x\frac{dy}{dx})=0\\ \frac{dy}{dx}(2\sin y \cos y-x\sin xy)= y\sin xy\\ \frac{dy}{dx} = \frac{y\sin xy}{2\sin y \cos y-x\sin xy} = \frac{y\sin xy}{\sin 2y -x\sin xy} \ \ \ \ \ \ (\because 2\sin x\cos y = \sin 2x)
Therefore, the answer is  \frac{y\sin xy}{\sin 2y -x\sin xy}

Posted by

Gautam harsolia

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