If f(x)=log _{x^{2}}left(log _{e} xright),the | vedclass

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(C) Given,$f(x) = \log _{x^{2}}(\log _{e} x) = \frac{1}{2} \log _{x}(\log _{e} x)$.Using the change of base formula,$f(x) = \frac{1}{2} \frac{\log _{e

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$\frac{1}{2e}$

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(C) Given,$f(x) = \log _{x^{2}}(\log _{e} x) = \frac{1}{2} \log _{x}(\log _{e} x)$.Using the change of base formula,$f(x) = \frac{1}{2} \frac{\log _{e}(\log _{e} x)}{\log _{e} x}$.Applying the quotient rule,$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\log _{e} x \cdot \frac{d}{dx}(\log _{e}(\log _{e} x)) - \log _{e}(\log _{e} x) \cdot \frac{d}{dx}(\log _{e} x)}{(\log _{e} x)^{2}} \right]$.$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\log _{e} x \cdot (\frac{1}{\log _{e} x} \cdot \frac{1}{x}) - \log _{e}(\log _{e} x) \cdot \frac{1}{x}}{(\log _{e} x)^{2}} \right]$.$f^{\prime}(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{1}{x} \log _{e}(\log _{e} x)}{(\log _{e} x)^{2}} \right]$.At $x = e$,$\log _{e} e = 1$ and $\log _{e}(\log _{e} e) = \log _{e} 1 = 0$.Substituting these values,$f^{\prime}(e) = \frac{1}{2} \left[ \frac{\frac{1}{e} - \frac{1}{e} \cdot 0}{(1)^{2}} \right] = \frac{1}{2e}$.

If $f(x)=\log _{x^{2}}\left(\log _{e} x\right)$,then $f^{\prime}(x)$ at $x=e$ is

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