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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Aicharya · Accepted Answer

$\begin{aligned} & f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}ight)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight) \ & \Rightarrow f(x)=\sin ^{-1}\left(\frac{2 	an 	heta}{1+	an ^2 	heta}ight)+\cos ^{-1}\left(\frac{1-	an ^2 	heta}{1+	an ^2 	heta}ight) \ & \Rightarrow f(x)=\sin ^{-1}(\sin 2 	heta)+\cos ^{-1}(\cos 2 	heta) \ & \Rightarrow f(x)=\pi-2 	heta+2 	heta=\pi \ & \Rightarrow f^{\prime}(x)=0\end{aligned}$

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,…

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