Question

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Answer 1

Step-by-step explanation:

answer above alternative also attached :)

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Answer 2

$\large\underline{\sf{Given \:Question - }}$

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

$\green{\large\underline{\sf{Solution-}}}$

Given expression is

$\rm :\longmapsto\: y = \dfrac{ {(1 - x)}^{2} }{ {x}^{2} }$

can be rewritten as

$\rm :\longmapsto\:y = {\bigg[\dfrac{1 - x}{x} \bigg]}^{2}$

$\rm :\longmapsto\:y = {\bigg[\dfrac{1}{x} - 1 \bigg]}^{2}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {\bigg[\dfrac{1}{x} - 1 \bigg]}^{2}$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2\bigg[\dfrac{1}{x} - 1\bigg] \: \dfrac{d}{dx}\bigg[\dfrac{1}{x} - 1\bigg]$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}k \: = \: 0 \: }}$

and

$\boxed{ \tt{ \: \dfrac{d}{dx} \frac{1}{ {x}^{n} } \: = \: \frac{ - n}{ {x}^{n + 1} } \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2\bigg[\dfrac{1}{x} - 1\bigg] \: \times \bigg[\dfrac{ - 1}{ {x}^{2} } - 0\bigg]$

$\rm \implies\:\dfrac{dy}{dx} = \: - \: \dfrac{2}{ {x}^{3} } + \dfrac{2}{ {x}^{2} }$

Hence,

$\red{ \boxed{\sf \: If \: y = \dfrac{ {(1 - x)}^{2} }{ {x}^{2} }, \: then \: \dfrac{dy}{dx} \: is \: - \: \dfrac{2}{ {x}^{3} } + \dfrac{2}{ {x}^{2} }}}$

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More to Know :-

$\green{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x}\\ \\ \sf {sin}^{ - 1}x & \sf \dfrac{1}{ \sqrt{1 - {x}^{2} } } \\ \\ \sf {cos}^{ - 1}x & \sf \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } } \\ \\ \sf {tan}^{ - 1}x & \sf \dfrac{1}{1 + {x}^{2} } \\ \\ \sf {sec}^{ - 1}x & \sf \dfrac{1}{x \sqrt{ {x}^{2} - 1 } } \end{array}} \\ \end{gathered}}$