If f(x) = log_{x^2} (ln x),then f'(e) is equa | vedclass

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(C) Given $f(x) = \log_{x^2} (\ln x)$.Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$,we have:$f(x) = \frac{\ln(\ln x)}{\ln x^2} = \

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$\frac{1}{2e}$

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(C) Given $f(x) = \log_{x^2} (\ln x)$.Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$,we have:$f(x) = \frac{\ln(\ln x)}{\ln x^2} = \frac{\ln(\ln x)}{2 \ln x}$.Now,differentiate with respect to $x$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:$f'(x) = \frac{1}{2} \left[ \frac{(\frac{1}{x \ln x})(\ln x) - (\ln(\ln x))(\frac{1}{x})}{(\ln x)^2} \right]$.$f'(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{\ln(\ln x)}{x}}{(\ln x)^2} \right] = \frac{1 - \ln(\ln x)}{2x(\ln x)^2}$.At $x = e$,$\ln x = 1$ and $\ln(\ln x) = \ln(1) = 0$.$f'(e) = \frac{1 - 0}{2e(1)^2} = \frac{1}{2e}$.

List-$I$	List-$II$
$A$. $\sec^{-1} x$	$I$. $\frac{1}{1-x^2}, x \in (-1, 1)$
$B$. $\tanh^{-1} x$	$II$. $\frac{-1}{\|x\| \sqrt{x^2+1}}, x \neq 0$
$C$. $\coth^{-1} x$	$III$. $\frac{1}{\|x\| \sqrt{x^2-1}}, \|x\| > 1$
$D$. $\operatorname{cosech}^{-1} x$	$IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$
	$V$. $\frac{-1}{\|x\| \sqrt{1-x^2}}, \|x\| < 1, x \neq 0$

If $f(x) = \log_{x^2} (\ln x)$,then $f'(e)$ is equal to . . . . . .

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