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