If phi(x) = log_{5} log_{3} x,then phi'(e) is | vedclass

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(B) Given $\phi(x) = \log_{5} (\log_{3} x)$.Using the change of base formula $\log_{a} b = \frac{\ln b}{\ln a}$,we can write:$\phi(x) = \frac{\ln(\log

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

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(B) Given $\phi(x) = \log_{5} (\log_{3} x)$.Using the change of base formula $\log_{a} b = \frac{\ln b}{\ln a}$,we can write:$\phi(x) = \frac{\ln(\log_{3} x)}{\ln 5} = \frac{\ln(\frac{\ln x}{\ln 3})}{\ln 5} = \frac{\ln(\ln x) - \ln(\ln 3)}{\ln 5}$.Differentiating with respect to $x$:$\phi'(x) = \frac{1}{\ln 5} \cdot \frac{d}{dx} (\ln(\ln x) - \ln(\ln 3))$.$\phi'(x) = \frac{1}{\ln 5} \cdot \frac{1}{\ln x} \cdot \frac{1}{x}$.Now,evaluating at $x = e$:$\phi'(e) = \frac{1}{\ln 5} \cdot \frac{1}{\ln e} \cdot \frac{1}{e}$.Since $\ln e = 1$,we get:$\phi'(e) = \frac{1}{e \ln 5}$.

If $\phi(x) = \log_{5} \log_{3} x$,then $\phi'(e)$ is equal to

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