Find frac{dx}{dy} for the equation sin^{2} y | vedclass

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Question

(A) The given relationship is $\sin^{2} y + \cos(xy) = \pi$.Differentiating both sides with respect to $y$,we get:$\frac{d}{dy}(\sin^{2} y) + \frac{d}

Vedclass · Accepted Answer

$\frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}$

Vedclass · Answer

(A) The given relationship is $\sin^{2} y + \cos(xy) = \pi$.Differentiating both sides with respect to $y$,we get:$\frac{d}{dy}(\sin^{2} y) + \frac{d}{dy}(\cos(xy)) = \frac{d}{dy}(\pi)$Using the chain rule:$\frac{d}{dy}(\sin^{2} y) = 2 \sin y \cos y \frac{dy}{dy} = \sin(2y)$.For the second term:$\frac{d}{dy}(\cos(xy)) = -\sin(xy) \cdot \frac{d}{dy}(xy) = -\sin(xy) \cdot (x + y \frac{dx}{dy})$.Substituting these into the equation:$\sin(2y) - \sin(xy) \cdot (x + y \frac{dx}{dy}) = 0$.Rearranging to solve for $\frac{dx}{dy}$:$\sin(2y) - x \sin(xy) - y \sin(xy) \frac{dx}{dy} = 0$.$y \sin(xy) \frac{dx}{dy} = \sin(2y) - x \sin(xy)$.Therefore,$\frac{dx}{dy} = \frac{\sin(2y) - x \sin(xy)}{y \sin(xy)}$.

Find $\frac{dx}{dy}$ for the equation $\sin^{2} y + \cos(xy) = \pi$.

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