Put x=tanθ⇒θ=tan−1x∴2x1+x2=2tanθ1+tan2θ=sin2θ∴ddxsin−1(sin2θ)=ddx(2θ)=ddx(2⋅tan−1x)=2⋅11+x2

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

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$\frac{2}{1 - x^{2}}$

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

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

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

answer is C.

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Detailed Solution

Put $x = \tan θ \Rightarrow θ = \tan^{- 1} x$

$∴ \frac{2 x}{1 + x^{2}} = \frac{2 \tan θ}{1 + \tan^{2} θ} = \sin 2 θ$

$∴ \frac{d}{d x} \sin^{- 1} (\sin 2 θ) = \frac{d}{d x} (2 θ)$

$= \frac{d}{d x} (2 \cdot \tan^{- 1} x)$

$= 2 \cdot \frac{1}{1 + x^{2}}$

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