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