If f(x) = log_{5} log_{3} x, then f^{prime}(e | vedclass

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(C) Given $f(x) = \log_{5} \log_{3} x$. Using the change of base formula $\log_{a} b = \frac{\ln b}{\ln a}$, we have: $f(x) = \frac{\ln(\log_{3} x)}{\

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$\frac{1}{e \log_{e} 5}$

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(C) Given $f(x) = \log_{5} \log_{3} x$. Using the change of base formula $\log_{a} b = \frac{\ln b}{\ln a}$, we have: $f(x) = \frac{\ln(\log_{3} x)}{\ln 5} = \frac{\ln(\frac{\ln x}{\ln 3})}{\ln 5} = \frac{\ln(\ln x) - \ln(\ln 3)}{\ln 5}$. Differentiating with respect to $x$: $f^{\prime}(x) = \frac{1}{\ln 5} \cdot \frac{d}{dx} [\ln(\ln x) - \ln(\ln 3)] = \frac{1}{\ln 5} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}$. Now, substituting $x = e$: $f^{\prime}(e) = \frac{1}{\ln 5} \cdot \frac{1}{\ln e} \cdot \frac{1}{e} = \frac{1}{\ln 5 \cdot 1 \cdot e} = \frac{1}{e \ln 5}$. Since $\ln 5 = \log_{e} 5$, we get $f^{\prime}(e) = \frac{1}{e \log_{e} 5}$.

If $f(x) = \log_{5} \log_{3} x$, then $f^{\prime}(e)$ is equal to

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