If f(x) = sin^{-1}left(frac{2x}{1+x^2}right) | vedclass

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(B) Let $x = 	an 	heta$. Since $x \in (1, \infty)$,we have $	heta \in (\frac{\pi}{4}, \frac{\pi}{2})$.Then $2	heta \in (\frac{\pi}{2}, \pi)$.$f(x)

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(B) Let $x = 	an 	heta$. Since $x \in (1, \infty)$,we have $	heta \in (\frac{\pi}{4}, \frac{\pi}{2})$.Then $2	heta \in (\frac{\pi}{2}, \pi)$.$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)$$f(x) = \sin^{-1}(\sin 2	heta) + \cos^{-1}(\cos 2	heta)$Since $2	heta \in (\frac{\pi}{2}, \pi)$,$\sin^{-1}(\sin 2	heta) = \pi - 2	heta$ and $\cos^{-1}(\cos 2	heta) = 2	heta$.Thus,$f(x) = (\pi - 2	heta) + 2	heta = \pi$.Since $f(x) = \pi$ is a constant function,its derivative $f'(x) = 0$.

If $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $x \in (1, \infty)$,then $f'(x)$ is equal to:

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