Question

If $ y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+........\infty ,}}} $ then $ \frac{dy}{dx} $ is equal to:

• A. $ \frac{\cos x}{2y-1} $
• B. $ \frac{-\cos x}{2y-1} $
• C. $ \frac{\sin x}{1-2y} $
• D. $ \frac{-\sin x}{1-2y} $
• E. $ \frac{2\cos x}{2y-1} $

Answer 1

Answer: (A)

$ y=\sqrt{\sin x+\sqrt{\sin x+....\infty }} $
$ \Rightarrow $ $ y=\sqrt{\sin x+y} $
$ \Rightarrow $ $ {{y}^{2}}=\sin x+y $
On differentiating w.r.t. $ x, $ we get
$ 2y\frac{dy}{dx}=\cos x+\frac{dy}{dx} $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{\cos x}{2y-1} $