Question

If $y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+...........\infty }}}$, then show that $\frac{dy}{dx}=\frac{\cos x}{2y-1}$

Answer 1

Given, $y=\sqrt{\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x\sqrt{\sin x+\sqrt{\sin x+........\infty }}]}}}}$
Thus, $y=\sqrt{\sin x+y}$
Squaring both sides, we get,
$y^2=\sin x+y$
Differentiating both sides, we get
$2y\frac{dy}{dx}=\frac{dy}{dx}+\cos x$
$\frac{dy}{dx}=\frac{\cos x}{(2y-1)}$