If $y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.... ~to~\infty}}}$, then value of $\frac{d y}{d x}$ is :

$\sqrt{\frac{\sin x}{y+1}}$ $\frac{\sin x}{y+1}$ $\frac{\cos x}{2 y+1}$ $\frac{\cos x}{2 y-1}$

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

$y=\sqrt{\sin x+y} \Rightarrow y^2=\sin x+y$ $\Rightarrow 2 y \frac{d y}{d x}=\cos x+\frac{d y}{d x} \Rightarrow(2 y-1) \frac{d y}{d x}=\cos x$ $\Rightarrow \frac{d y}{d x}=\frac{\cos x}{2 y-1}$ Hence (4) is correct answer.