If y=sin−11−x21+x2, then dydx=

y=sin−11−x21+x2Let x=tanθy=sin−1(1−tan2θ1+tan2θ)=sin−1(cos2θ)⇒sin−1(sin(π2−2θ))=π2−2θy=π2−2tan−1xdydx=−21+x2

If y=sin−11−x21+x2, then dydx=

y=sin−11−x21+x2Let x=tanθy=sin−1(1−tan2θ1+tan2θ)=sin−1(cos2θ)⇒sin−1(sin(π2−2θ))=π2−2θy=π2−2tan−1xdydx=−21+x2

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If $y = \sin^{- 1} \frac{1 - x^{2}}{1 + x^{2}}$ , then $\frac{d y}{d x} =$

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

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

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

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

answer is A.

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Detailed Solution

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

$Let x = \tan θ$

$y = \sin^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) = \sin^{- 1} (\cos 2 θ)$

$\Rightarrow \sin^{- 1} (\sin (\frac{π}{2} - 2 θ)) = \frac{π}{2} - 2 θ$

$y = \frac{π}{2} - 2 \tan^{- 1} x$

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

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If $y = \sin^{- 1} \frac{1 - x^{2}}{1 + x^{2}}$ , then $\frac{d y}{d x} =$