Explanation

(a)
$y = \sqrt{\sin{x + \sqrt{\sin{x + \sqrt{\sin{x + ...\infty}}}}}}$
$\Rightarrow y = \sqrt{\sin{x + y}}$
$\Rightarrow y^{2} = \sin{x + y}$
On differentiating w.r.t. $x$, we get
$2y\frac{dy}{dx} = \cos{x + \frac{dy}{dx}}$
$\Rightarrow \frac{dy}{dx} = \frac{\cos x}{2y - 1}$