Q. If $y = \sqrt{x + \sqrt{x + \sqrt{x + ...to\ \infty,}}}$ then $\frac{dy}{dx}$ is equal to
A) $\frac{x}{2y - 1}$
B) $\frac{2}{2y - 1}$
C) $- \frac{1}{2y - 1}$
D) $\frac{1}{2y - 1}$
Explanation
(d)
$y = \sqrt{x + \sqrt{x + \sqrt{x + ...\infty}}}$
$\Rightarrow y = \sqrt{x + y}$
On squaring both sides, we get
$y^{2} = x + y\ \Rightarrow \ 2y\frac{dy}{dx} = 1 + \frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}(2y - 1) = 1\ \Rightarrow \ \frac{dy}{dx} = \frac{1}{2y - 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.