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

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(A) Given $f(x) = \cos^{-1} \left( \frac{1 - (\log x)^2}{1 + (\log x)^2} \right)$.Let $u = \log x$. Then the expression becomes $f(x) = \cos^{-1} \lef

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$1/e$

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(A) Given $f(x) = \cos^{-1} \left( \frac{1 - (\log x)^2}{1 + (\log x)^2} ight)$.Let $u = \log x$. Then the expression becomes $f(x) = \cos^{-1} \left( \frac{1 - u^2}{1 + u^2} ight)$.Using the trigonometric substitution $u = 	an 	heta$,we know that $\cos^{-1} \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight) = \cos^{-1} (\cos 2	heta) = 2	heta = 2 	an^{-1} u$.Thus,$f(x) = 2 	an^{-1} (\log x)$.Differentiating with respect to $x$ using the chain rule:$f'(x) = 2 \cdot \frac{1}{1 + (\log x)^2} \cdot \frac{d}{dx}(\log x) = \frac{2}{1 + (\log x)^2} \cdot \frac{1}{x}$.Now,evaluating at $x = e$:$f'(e) = \frac{2}{1 + (\log e)^2} \cdot \frac{1}{e} = \frac{2}{1 + 1^2} \cdot \frac{1}{e} = \frac{2}{2} \cdot \frac{1}{e} = \frac{1}{e}$.

If $f(x) = \cos^{-1} \left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right]$,then $f'(e) = \_\_\_\_$

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