If y = sqrt{sin x + y}, then frac{dy}{dx} is | vedclass

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(B) Given the equation $y = \sqrt{\sin x + y}.$Squaring both sides,we get $y^2 = \sin x + y.$Differentiating both sides with respect to $x$,we have:$\

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$\frac{\cos x}{2y - 1}$

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(B) Given the equation $y = \sqrt{\sin x + y}.$Squaring both sides,we get $y^2 = \sin x + y.$Differentiating both sides with respect to $x$,we have:$\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x + y)$$2y \cdot \frac{dy}{dx} = \cos x + \frac{dy}{dx}$Rearranging the terms to isolate $\frac{dy}{dx}$:$2y \cdot \frac{dy}{dx} - \frac{dy}{dx} = \cos x$$\frac{dy}{dx}(2y - 1) = \cos x$Therefore,$\frac{dy}{dx} = \frac{\cos x}{2y - 1}.$

If $y = \sqrt{\sin x + y},$ then $\frac{dy}{dx}$ is equal to

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