Question

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

Answer 1

Given, $y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+.....\infty }}}$

$\Rightarrow y=\sqrt{\cos x+y}$ $[$Given $y=\sqrt{\cos x+\sqrt{\cos x+......\infty }}]$

Squaring both the sides we get $y^2=\cos x+y$

Differentiating both the sides w.r.t. $x$ we get.

$\frac{d}{dx}{y}^2=\frac{d}{dx}\cos x+\frac{dy}{dx}$

$\Rightarrow 2y\frac{dy}{dx}=-\sin x+\frac{dy}{dx}$

$\Rightarrow 2y\frac{dy}{dx}-\frac{dy}{dx}=-\sin x$

$\Rightarrow (2y-1)\frac{dy}{dx}=-\sin x$

$\Rightarrow \frac{dy}{dx}=-\frac{\sin x}{(2y-1)}$.