If cos (x + y) = ysin x, then frac{dy}{dx} = | vedclass

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Question

(A) Given the equation: $\cos (x + y) = y\sin x$ Differentiating both sides with respect to $x$ using the chain rule and product rule: $-\sin (x + y)

Vedclass · Accepted Answer

$ - \frac{\sin (x + y) + y\cos x}{\sin x + \sin (x + y)}$

Vedclass · Answer

(A) Given the equation: $\cos (x + y) = y\sin x$ Differentiating both sides with respect to $x$ using the chain rule and product rule: $-\sin (x + y) \cdot \frac{d}{dx}(x + y) = y \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{dy}{dx}$ $-\sin (x + y) \left( 1 + \frac{dy}{dx} \right) = y\cos x + \sin x \frac{dy}{dx}$ Expanding the left side: $-\sin (x + y) - \sin (x + y) \frac{dy}{dx} = y\cos x + \sin x \frac{dy}{dx}$ Grouping the $\frac{dy}{dx}$ terms on one side: $-\sin (x + y) - y\cos x = \sin x \frac{dy}{dx} + \sin (x + y) \frac{dy}{dx}$ $-(\sin (x + y) + y\cos x) = \frac{dy}{dx} (\sin x + \sin (x + y))$ Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = - \frac{\sin (x + y) + y\cos x}{\sin x + \sin (x + y)}$ Thus,the correct option is $A$.

If $\cos (x + y) = y\sin x,$ then $\frac{dy}{dx} = $

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