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Question: If \(y = \sqrt{x + \sqrt{x + \sqrt{x + ........\text{to }\infty}}}\) then \(\frac{dy}{dx} =\)...

If y=x+x+x+........to ∞y = \sqrt{x + \sqrt{x + \sqrt{x + ........\text{to }\infty}}} then dydx=\frac{dy}{dx} =

A

x2y−1\frac{x}{2y - 1}

B

22y−1\frac{2}{2y - 1}

C

−12y−1\frac{- 1}{2y - 1}

D

−12y−1\frac{- 1}{2y - 1}

Answer

−12y−1\frac{- 1}{2y - 1}

Explanation

Solution

y=x+x+x+........to ∞y = \sqrt{x + \sqrt{x + \sqrt{x + ........\text{to }\infty}}} ⇒ y=x+yy = \sqrt{x + y} ⇒ y2=x+yy^{2} = x + y

⇒ 2ydydx=1+dydx2y\frac{dy}{dx} = 1 + \frac{dy}{dx} ⇒ dydx(2y−1)=1\frac{dy}{dx}(2y - 1) = 1 ⇒ dydx=12y−1\frac{dy}{dx} = \frac{1}{2y - 1}