ddxsin−1⁡2x−1+x is equal to

A
$\frac{2}{1 + x^{2}}, 0 < | x | < 1$
B
$\frac{2}{\sqrt{1 - x^{2}}}, | x | < 1$
C
$\frac{2}{\sqrt{1 + x^{2}}}$
D
$0$

Solution:

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

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

Solution:

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