If f(x) = log_{x^2}(log x),then at x = e,f'(x | vedclass

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

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$\frac{1}{2e}$

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

If $f(x) = \log_{x^2}(\log x)$,then at $x = e$,$f'(x)$ has the value

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