Question:

<b>Derivative of Iterated Logarithm</b><br> $f(x) = \dfrac{\log_e(\log_e x)}{\log_e x} = \dfrac{\ln(\ln x)}{\ln x}$.<br> Using quotient rule, let $u=\ln(\ln x)$ and $v=\ln x$:<br> $u'=\dfrac{1}{\ln x}\cdot\dfrac{1}{x}=\dfrac{1}{x\ln x}$, $v'=\dfrac{1}{x}$.<br> $f'(x)=\dfrac{u'v-uv'}{v^2}=\dfrac{\frac{1}{x\ln x}\cdot\ln x - \ln(\ln x)\cdot\frac{1}{x}}{(\ln x)^2}$<br> $=\dfrac{\frac{1}{x}-\frac{\ln(\ln x)}{x}}{(\ln x)^2}=\dfrac{1-\ln(\ln x)}{x(\ln x)^2}$.<br> At $x=e$: $\ln e=1$, $\ln(\ln e)=\ln 1=0$.<br> $f'(e)=\dfrac{1-0}{e\cdot 1}=\dfrac{1}{e}$.<br> If the question asks for $e\cdot f'(e)=1$, <b>Answer = 1</b>.<br> <b>Key concept:</b> $\log_x(\log_e x)=\ln(\ln x)/\ln x$; apply quotient rule.<br> <b>Trap:</b> Confusing $\log_x$ with $\log_e$; remember $\log_x y = \ln y/\ln x$.