Question

find the derivative of sin^(-1)+(x^2-1/x^2+1)​

Answer 1

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

Given function is

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

Let assume that,

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

We know,

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

So, using this identity, we get

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

can be further rewritten as

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

We know that,

$\boxed{ \tt{ \: {cos}^{ - 1}( - x) = \pi \: - \: {cos}^{ - 1}x \: }}$

So, using this identity, we get

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

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

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

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

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

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

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

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

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