If f(x) = log_{x}(log x),then f'(x) at x = e | vedclass

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(B) Given $f(x) = \log_{x}(\log x)$.Using the base change formula,we can write $f(x) = \frac{\log(\log x)}{\log x}$.Now,differentiate $f(x)$ with resp

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$\frac{1}{e}$

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(B) Given $f(x) = \log_{x}(\log x)$.Using the base change formula,we can write $f(x) = \frac{\log(\log x)}{\log x}$.Now,differentiate $f(x)$ with respect to $x$ using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:$f'(x) = \frac{\left(\frac{d}{dx}(\log(\log x))\right)(\log x) - (\log(\log x))(\frac{d}{dx}(\log x))}{(\log x)^2}$$f'(x) = \frac{\left(\frac{1}{\log x} \cdot \frac{1}{x}\right)(\log x) - (\log(\log x))(\frac{1}{x})}{(\log x)^2}$$f'(x) = \frac{\frac{1}{x} - \frac{\log(\log x)}{x}}{(\log x)^2} = \frac{1 - \log(\log x)}{x(\log x)^2}$.Now,evaluate $f'(x)$ at $x = e$:$f'(e) = \frac{1 - \log(\log e)}{e(\log e)^2}$.Since $\log e = 1$ and $\log(1) = 0$,we have:$f'(e) = \frac{1 - 0}{e(1)^2} = \frac{1}{e}$.

If $f(x) = \log_{x}(\log x)$,then $f'(x)$ at $x = e$ is

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