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

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

The correct answer is Option (1) → $\frac{\cos x}{2y - 1}$ ##

Since, $y = (\sin x + y)^{1/2}$

$∴\frac{dy}{dx} = \frac{1}{2}(\sin x + y)^{-1/2} \cdot \frac{d}{dx}(\sin x + y)$

$⇒ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{(\sin x + y)^{1/2}} \cdot \left( \cos x + \frac{dy}{dx} \right)$

$⇒ \frac{dy}{dx} = \frac{1}{2y} \left( \cos x + \frac{dy}{dx} \right) \quad [∵(\sin x + y)^{1/2} = y]$

$⇒\frac{dy}{dx} \left( 1 - \frac{1}{2y} \right) = \frac{\cos x}{2y}$

$∴\frac{dy}{dx} = \frac{\cos x}{2y} \cdot \frac{2y}{2y - 1} = \frac{\cos x}{2y - 1}$