Q. If $y = 1 + \frac{1}{x}\ + \frac{1}{x^{2}} + \frac{1}{x^{3}} + \ldots\infty$ with $|x| > 1,\ then\frac{dy}{dx}\ is\$
A) $\frac{x^{2}}{y^{2}}$
B) $x^{2}y^{2}$
C) $\frac{y^{2}}{x^{2}}$
D) $- \frac{y^{2}}{x^{2}}$
Explanation
(d)
Given, $y = 1 + \frac{1}{x}$+$\frac{1}{x^{2}} + \frac{1}{x^{3}} + .\ldots = \frac{1}{1 - \frac{1}{x}}$
$\Rightarrow \ \ y = \frac{x}{x - 1}$(GP series) …(i)
$\frac{dy}{dx} =$ $\frac{1(x - 1) - x.1}{(x - 1)^{2}} = - \frac{1}{(x - 1)^{2}}$
$\Rightarrow \ \ \ \frac{dy}{dx} = \ - \frac{y^{2}}{x^{2}}\ \ \ \ \$ [from Eq.(i)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.