Question

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

• A. $ \frac{1}{2y+1} $
• B. $ \frac{1}{2y-1} $
• C. $ \frac{1}{xy} $
• D. 1

Answer 1

Answer: (B)

Let $ y=\sqrt{x+\sqrt{x+\sqrt{x+....\infty }}} $
$ \Rightarrow $ $ y=\sqrt{x+y} $
$ \Rightarrow $ $ {{y}^{2}}=x+y $
$ \Rightarrow $ $ {{y}^{2}}-y-x=0 $
On differentiating w.r.t. $ x, $ we get
$ 2y\frac{dy}{dx}-\frac{dy}{dx}-1=0 $
$ \Rightarrow $ $ \frac{dy}{dx}=\frac{1}{(2y-1)} $