If $y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then find $\frac{dy}{dx}$

If $y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then find $\frac{dy}{dx}$
  1. \(\frac{d y}{d x}=\frac{-2}{1+x^2}\)
  2. \(\frac{d y}{d x}=\frac{-x}{1+x^2}\)
  3. \(\frac{d y}{d x}=\frac{2}{1+x^2}\)
  4. \(\frac{d y}{d x}=\frac{\sqrt{1}}{1+x^2}\)

Solution

\(\begin{aligned}
& y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\frac{\pi}{2}-\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\
& \text {Put } x=\tan \theta \Rightarrow y=\frac{\pi}{2}-\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\
& \Rightarrow y=\frac{\pi}{2}-\cos ^{-1} \cos (2 \theta)=\frac{\pi}{2}-2 \theta \\
& \Rightarrow y=\frac{\pi}{2}-2 \tan ^{-1} x \therefore \frac{d y}{d x}=-\frac{2}{1+x^2}
\end{aligned}\)

Asked in: MHT CET 2020 (19 Oct Shift 2)

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