If y = frac{(1 - x)^2}{x^2},then frac{dy}{dx} | vedclass

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(D) Given $y = \frac{(1 - x)^2}{x^2}$.Expanding the numerator,we get $y = \frac{1 - 2x + x^2}{x^2}$.Dividing each term by $x^2$,we get $y = \frac{1}{x

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

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(D) Given $y = \frac{(1 - x)^2}{x^2}$.Expanding the numerator,we get $y = \frac{1 - 2x + x^2}{x^2}$.Dividing each term by $x^2$,we get $y = \frac{1}{x^2} - \frac{2x}{x^2} + \frac{x^2}{x^2} = x^{-2} - 2x^{-1} + 1$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(x^{-2}) - 2\frac{d}{dx}(x^{-1}) + \frac{d}{dx}(1)$.Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$,we get $\frac{dy}{dx} = -2x^{-3} - 2(-1)x^{-2} + 0$.$\frac{dy}{dx} = -\frac{2}{x^3} + \frac{2}{x^2}$.Thus,the correct option is $D$.

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

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