Question

If $y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+}....\infty }}$ then find the value of $\frac{dy}{dx}$.

Answer 1

$y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\dots \infty }}}$

Squaring both sides, we have

$y^2=\cos x+\sqrt{\cos x+\sqrt{\cos x+\dots \infty }}$

But, as $y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\dots \infty }}}$, we have

$y^2=\cos x+y$

$⟹y^2-y=\cos x$

On differentiating wrt $x$,

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

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

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

$\frac{dy}{dx}=\frac{\sin x}{1-2\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\dots \infty }}}}$

This is the required solution.