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If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right), x \in(1, \infty)$, then $f^{\prime}(x)$
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$\begin{aligned} & f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\ & \Rightarrow f(x)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\ & \Rightarrow f(x)=\sin ^{-1}(\sin 2 \theta)+\cos ^{-1}(\cos 2 \theta) \\ & \Rightarrow f(x)=\pi-2 \theta+2 \theta=\pi \\ & \Rightarrow f^{\prime}(x)=0\end{aligned}$
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