Let $y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....}}}$, then $\frac{dy}{dx}$ is:
Answer & explanation
Correct answer: option 2
$y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....}}}=\sqrt{\sin x+y}$
$⇒y^2=\sin x+y⇒2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}$
$⇒\frac{dy}{dx}=\frac{\cos x}{2y-1}$