Question

Differentiation

$if \: y = {cos}^{ - 1} \binom{2x}{1 + {x}^{2} } \: then \: \frac{dy}{dx} =$
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Answer 1

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

Given function is

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

We know,

$\boxed{ \tt{ \: {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2} \: }}$

So, using this identity, we get

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

We know,

$\boxed{ \tt{ \: {sin}^{ - 1}\bigg[\dfrac{2x}{1 + {x}^{2} } \bigg] = 2 {tan}^{ - 1}x \: }}$

So, using this identity, we get

$\rm :\longmapsto\:y = \dfrac{\pi}{2} - {2tan}^{ - 1}x$

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

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

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}\dfrac{\pi}{2} - \dfrac{d}{dx} {2tan}^{ - 1}x$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}k \: = \: 0 \: }} \\ \\ and \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx} {tan}^{1}x = \frac{1}{1 + {x}^{2} } \: }}$

So, using this result, we get

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

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} = \frac{ - 2}{1 + {x}^{2} } \: }}$

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More to know :-

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