Q. For $|x| < 1$, let$\ y = 1 + x + x^{2}\ldots$to $\infty$, then $\frac{dy}{dx}$ equal to
A) $\frac{x}{y}$
B) $\frac{x^{2}}{y^{2}}$
C) $\frac{x}{y^{2}}$
D) $xy^{2} + y$
Explanation
(d)
$\because y = 1 + x + x^{2} + ...\infty$
$\therefore y = \frac{1}{1 - x} = (1 - x)^{- 1}$
On differentiating w.r.t. $x,$ we get
$\frac{dy}{dx} = - \frac{1}{(1 - x)^{2}}( - 1) = \frac{1}{(1 - x)^{2}}$
$\therefore\ \frac{dy}{dx} - y = \frac{1}{(1 - x)^{2}} - \frac{1}{(1 - x)}$
$= \frac{1 - 1 + x}{(1 - x)^{2}} = \frac{x}{(1 - x)^{2}}$
$\Rightarrow \ \frac{dy}{dx} - y = xy^{2}$
$\Rightarrow \ \frac{dy}{dx} = xy^{2} + y$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.