Question

If $y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+}}}...\infty$ then $\frac{dy}{dx}=$

Answer 1

$y=\sqrt{sinx+\sqrt{sinx+\sqrt{sinx+............\infty }}}$

Let $y=\sqrt{sinx+y}$

$y^2=sinx+y$

differentiat w.r.t.$x$,

$2y.\frac{dy}{dx}=cosx+\frac{dy}{dx}$

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