Question

If $y=1+\frac {1}{x} +\frac{1}{x^2}+\frac {1}{x^3}+.......... $ to $\infty $ with $ |x|>1$ then $\frac {dy}{dx}$=

• A. $\frac{-y^2}{x^2}$
• B. $\frac{y^2}{x^2}$
• C. $x^2 y^2$
• D. $\frac{x^2}{y^2}$

Answer 1

Answer: (A)

We have
$y=1+\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}+\ldots$
$y =\frac{1}{1-\frac{1}{x}} $ (GP series)
$=\frac{x}{x-1}\,\,\,\,\dots(i)$
On differentiating, we get
$\frac{d y}{d x}=\frac{(x-1)-x(+1)}{(1-x)^{2}}=-\frac{1}{(1-x)^{2}} $
$=-\frac{y^{2}}{x^{2}}$ (From (i))