If f(x)=log _{x^2}(log x),then f^{prime}(x) a | vedclass

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(D) $f(x) = \log _{x^2}(\log x)$ Using the change of base formula $\log _a b = \frac{\log b}{\log a}$,we have: $f(x) = \frac{\log(\log x)}{\log(x^2)}

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$(2 e)^{-1}$

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(D) $f(x) = \log _{x^2}(\log x)$ Using the change of base formula $\log _a b = \frac{\log b}{\log a}$,we have: $f(x) = \frac{\log(\log x)}{\log(x^2)} = \frac{\log(\log x)}{2 \log x}$ Now,differentiate with respect to $x$ using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$: $f'(x) = \frac{1}{2} \frac{d}{dx} \left( \frac{\log(\log x)}{\log x} \right)$ $f'(x) = \frac{1}{2} \left[ \frac{(\log x) \cdot \frac{d}{dx}(\log(\log x)) - \log(\log x) \cdot \frac{d}{dx}(\log x)}{(\log x)^2} \right]$ $f'(x) = \frac{1}{2} \left[ \frac{(\log x) \cdot \frac{1}{\log x} \cdot \frac{1}{x} - \log(\log x) \cdot \frac{1}{x}}{(\log x)^2} \right]$ $f'(x) = \frac{1}{2x} \left[ \frac{1 - \log(\log x)}{(\log x)^2} \right]$ At $x = e$,$\log x = \log e = 1$ and $\log(\log x) = \log(1) = 0$: $f'(e) = \frac{1}{2e} \left[ \frac{1 - 0}{(1)^2} \right] = \frac{1}{2e} = (2e)^{-1}$

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

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