Question

If $y=\frac{(1-x{)}^2}{x^2}wherex\ne 0,then\frac{dy}{dx}is$

• A. $\frac{2}{x^2}+\frac{2}{x^3}$
• B. $-\frac{2}{x^3}+\frac{2}{x^2}$
• C. $-\frac{2}{x^2}+\frac{2}{x^3}$
• D. $-\frac{2}{x^2}-\frac{2}{x^3}$

Answer 1

The correct option is B $-\frac{2}{x^3}+\frac{2}{x^2}$

$y=\frac{{(1-x)}^2}{x^2}$

We know,

$\frac{d}{dt}\left(\frac{u}{v}\right)$ $=\frac{u\frac{d}{dt}\left(v\right)-\left(v\right)\frac{d}{dt}\left(u\right)}{v^2}$

$\Rightarrow \frac{dy}{dx}=\frac{x^2\left(2(1-x)(-1)\right)-{(1-x)}^2\left(2x\right)}{x^4}$

$=\frac{x^2(2x-2)-(1+x^2-2x)\left(2x\right)}{x^4}$

$=\frac{2x^2-2x}{x^4}$

$=\frac{-2}{x^3}+\frac{2}{x^2}$