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

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(D) Let $u = \log x$. Then $f(x) = \sin^{-1}\left(\frac{2u}{1+u^2}ight)$.Using the substitution $u = 	an 	heta$,we have $\frac{2u}{1+u^2} = \sin(2

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$\frac{1}{e}$

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(D) Let $u = \log x$. Then $f(x) = \sin^{-1}\left(\frac{2u}{1+u^2}ight)$.Using the substitution $u = 	an 	heta$,we have $\frac{2u}{1+u^2} = \sin(2	heta)$.Thus,$f(x) = \sin^{-1}(\sin(2	heta)) = 2	heta = 2 	an^{-1}(u) = 2 	an^{-1}(\log x)$.Now,differentiate with respect to $x$:$f^{\prime}(x) = 2 	imes \frac{1}{1+(\log x)^2} 	imes \frac{d}{dx}(\log x)$$f^{\prime}(x) = \frac{2}{1+(\log x)^2} 	imes \frac{1}{x} = \frac{2}{x(1+(\log x)^2)}$.Substitute $x = e$:$f^{\prime}(e) = \frac{2}{e(1+(\log e)^2)} = \frac{2}{e(1+1^2)} = \frac{2}{e(2)} = \frac{1}{e}$.

If $f(x) = \sin^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right)$,then $f^{\prime}(e)$ is

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