If $y =\sqrt{x+\sqrt{x+\sqrt{x+.....}}}$, then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $(2y - 1)\frac{dy}{dx} -1= 0$
Given: $y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}$
Let $y = \sqrt{x + y}$
Square both sides: $y^2 = x + y \Rightarrow y^2 - y - x = 0$
Differentiating both sides w.r.t $x$:
$2y \frac{dy}{dx} - \frac{dy}{dx} - 1 = 0 \Rightarrow (2y - 1)\frac{dy}{dx} = 1$