Derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ w.r.t. $\tan ^{-1} x,-1 < x < 1$ is

Derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ w.r.t. $\tan ^{-1} x,-1 < x < 1$ is
  1. 2
  2. $\frac{1}{1+x^2}$
  3. $\frac{2}{1+x^2}$
  4. $\frac{1}{2}$

Solution

$\begin{aligned} & \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \text { let } x=\tan \theta \\ & =\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta=2 \tan ^{-1} x \\ & \text { Now } \frac{d\left(\sin ^{-1} \frac{2 x}{1+x^2}\right)}{d\left(\tan ^{-1} x\right)}=\frac{d\left(2 \tan ^{-1} x\right)}{d\left(\tan ^{-1} x\right)}=2\end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 2)

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