ddxsin−1⁡2x−1+x is equal to

y=sin−1⁡2x−1+x=sin−1⁡2x1+x2 put x= tanθ =sin−1⁡2tanθ1+tan2θ=sin−1⁡sin2θ=2θy=2tan-1xdydx=21+x2

ddxsin−1⁡2x−1+x is equal to

y=sin−1⁡2x−1+x=sin−1⁡2x1+x2 put x= tanθ =sin−1⁡2tanθ1+tan2θ=sin−1⁡sin2θ=2θy=2tan-1xdydx=21+x2

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$\frac{d}{d x} (\sin^{- 1} (\frac{2}{x^{- 1} + x}))$ is equal to

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$\frac{2}{1 + x^{2}}, 0 < | x | < 1$

$0$

$\frac{2}{\sqrt{1 - x^{2}}}, | x | < 1$

$\frac{2}{\sqrt{1 + x^{2}}}$

answer is A.

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Detailed Solution

$y = \sin^{- 1} (\frac{2}{x^{- 1} + x})$

$= \sin^{- 1} (\frac{2 x}{1 + x^{2}})$

$p u t x = \tan θ = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) = \sin^{- 1} (\sin 2 θ) = 2 θ$

$y = 2 \tan^{- 1} x$

$\frac{d y}{d x} = \frac{2}{1 + x^{2}}$

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$\frac{d}{d x} (\sin^{- 1} (\frac{2}{x^{- 1} + x}))$ is equal to