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

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

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$\frac{1}{2e}$

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

If $f(x) = \log_{x^2}(\log_{e} x)$,then $f^{\prime}(x)$ at $x = e$ is

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