Question

For $|x|<1,$ let $y=1+x+x^2+.....$ to $\infty ,$ then $\frac{dy}{dx}-y$ is equal to:

• A. $\frac{x}{y}$
• B. $\frac{x^2}{y^2}$
• C. $\frac{x}{y^2}$
• D. $x{y}^2$
• E. $\frac{x-1}{y^2}$

Answer 1

Answer: (C)

$\because$ $y=1+x+x^2+....\infty$ $\therefore$ $y=\frac{1}{1-x}={(1-x)}^{-1}$ $\Rightarrow$ $\frac{dy}{dx}=-\frac{1}{{(1-x)}^2}(-1)=\frac{1}{{(1-x)}^2}$ $\therefore$ $\frac{dy}{dx}-y=\frac{1}{{(1-x)}^2}-\frac{1}{1-x}=\frac{1-1+x}{{(1-x)}^2}$ $=\frac{x}{{(1-x)}^2}$ $\Rightarrow$ $\frac{dy}{dx}-y=\frac{x}{y^2}$