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If $f(x)=\log _{5} \log _{3} x,$ then $f^{\prime}(e)$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{e \log _{e} 5}$
We have,
$\therefore$
$\begin{aligned} f(x) &=\log _{5} \log _{3} x \\ &=\frac{\log \log _{3} x}{\log 5} \\ &=\frac{\log \left(\frac{\log x}{\log 3}\right)}{\log 5} \\ &=\frac{\log \log x-\log \log 3}{\log 5} \\ f^{\prime}(x)=\frac{1}{\log 5} \cdot \frac{1}{\log x} \cdot \frac{1}{x} \end{aligned}$
$\begin{aligned} f^{\prime}(e) &=\frac{1}{e \log 5 \cdot \log e} \\ &=\frac{1}{e \log 5} \quad[\because \log e=1] \\ &=\frac{1}{e \log _{n} 5} \end{aligned}$
$\therefore$
$\begin{aligned} f(x) &=\log _{5} \log _{3} x \\ &=\frac{\log \log _{3} x}{\log 5} \\ &=\frac{\log \left(\frac{\log x}{\log 3}\right)}{\log 5} \\ &=\frac{\log \log x-\log \log 3}{\log 5} \\ f^{\prime}(x)=\frac{1}{\log 5} \cdot \frac{1}{\log x} \cdot \frac{1}{x} \end{aligned}$
$\begin{aligned} f^{\prime}(e) &=\frac{1}{e \log 5 \cdot \log e} \\ &=\frac{1}{e \log 5} \quad[\because \log e=1] \\ &=\frac{1}{e \log _{n} 5} \end{aligned}$
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