Question

If $y=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\dots .\infty$ with $\left|x\right|>1$, then $\frac{dy}{dx}$ is equal to:

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

Answer 1

The correct option is D

$\frac{-y^2}{x^2}$

Explanation for the correct option.

Step 1: Simplify the given equation.

$1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\dots .\infty$ are in infinite G.P, where $a=1\mathrm{and}r=\frac{1}{x}$.

Sum of infinite terms of G.P $=\frac{a}{1-r}$.

So,

$\begin{array}{rcl}y& =& \frac{1}{1-{\displaystyle \frac{1}{x}}}\\ & =& \frac{x}{x-1}.....\left(1\right)\end{array}$

Step 2: Find $\frac{dy}{dx}$.

By differentiating w.r.t $x$, we get

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{\left(x-1\right)\left(1\right)-\left(x\right)\left(1\right)}{{\left(x-1\right)}^2}\\ & =& \frac{x-1-x}{{\left(x-1\right)}^2}\\ & =& \frac{-1}{{\left(x-1\right)}^2}\\ & =& \frac{-1}{{\left({\displaystyle \frac{x}{y}}\right)}^2}\left[\mathrm{from}\left(1\right)\right]\\ & =& \frac{-y^2}{x^2}\end{array}$

Hence, option D is correct.