Q. If $y =\frac{(1-x)^{2}}{x^{2}}$, then $\frac{d y}{d x}$

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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

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

Solution:

$y =\frac{(1-x)^{2}}{x^{2}}=\frac{1+x^{2}-2 x}{x^{2}}=\frac{1}{x^{2}}+1-\frac{2}{x}$
$\therefore \frac{d y}{d x}=-\frac{2}{x^{3}}+\frac{2}{x^{2}}$

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