Question

Find $\frac{dy}{dx}$ for the following
$y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+...\infty }}}$

Answer 1

$y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x}}+.......\infty }y=\sqrt{\cos x+y}{y}^2={\left(\sqrt{\cos x+y}\right)}^2\frac{d{y}^2}{dx}=\frac{dy}{dx}\left(\cos x+y\right)2y\frac{dy}{dx}=-\sin x+\frac{dy}{dx}2y\frac{dy}{dx}-\frac{dy}{dx}=-\sin x+\frac{dy}{dx}-\frac{dy}{dx}\frac{dy}{dx}\left(2y-1\right)=-\sin x\frac{dy}{dx}=\frac{-\sin x}{2y-1}$