If sin^2 x + 2cos y + xy = 0,then frac{dy}{dx | vedclass

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(B) Given equation: $\sin^2 x + 2\cos y + xy = 0$Differentiating both sides with respect to $x$:$\frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(2\cos y) + \fra

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$\frac{y + \sin 2x}{2\sin y - x}$

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(B) Given equation: $\sin^2 x + 2\cos y + xy = 0$Differentiating both sides with respect to $x$:$\frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(2\cos y) + \frac{d}{dx}(xy) = 0$Using the chain rule and product rule:$2\sin x \cos x - 2\sin y \frac{dy}{dx} + (y \cdot 1 + x \frac{dy}{dx}) = 0$Using the identity $\sin 2x = 2\sin x \cos x$:$\sin 2x - 2\sin y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx}(x - 2\sin y) = -(y + \sin 2x)$$\frac{dy}{dx} = \frac{y + \sin 2x}{2\sin y - x}$

If $\sin^2 x + 2\cos y + xy = 0$,then $\frac{dy}{dx} = $

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