Derivatives

If $f(x)=y$ : $y=\log_x(\ln~x)\\\\x^y=\ln x\\\\\ln(x^y)=\ln(\ln x)\\\\ y\cdot\ln x=\ln(\ln x)\\\\\\\\\\\\\\\\$

Derivating the expression: $y'\cdot\ln x+y\cdot\dfrac{1}{x}=\dfrac{1}{\ln x}\cdot\dfrac{1}{x}\\\\ y'\cdot\ln x+\dfrac{\log_x(\ln x)}{x}=\dfrac{1}{x\cdot\ln x}\\\\\\\\\\\\$

At $x=e$ : $y'\cdot\ln e+\dfrac{\log_e(\ln e)}{e}=\dfrac{1}{e\cdot\ln e}\\\\ y'\cdot1+\dfrac{\log_e(1)}{e}=\dfrac{1}{e\cdot1}\\\\ y'+\dfrac{0}{e}=\dfrac{1}{e}\\\\y'=e^{-1}\Longrightarrow\boxed{a=-1}$

f(x) = log(x) (lnx)

use this

f(x) = log(a) x

then f'(x) = 1 / x ln a

now asume lnx as p

let f(x) = y

then dy / dx = dy/dp . dp/dx

dy/dx = (1 / p. lnx ) . 1 / x

dy/dx = 1/ lnx . lnx . x

now subt x = e

then we get dy/dx = 1/ lne. lne . e = 1/e

so dy/dx = 1 / e = e^-1

then a = -1

$\begin{aligned} f(x) & = \log_x (\ln x) \\ & = \frac {\color{#3D99F6}{\ln (\ln x)}}{\color{#D61F06}{\ln x}} & \small \color{#3D99F6}{\text{By chain rule}} \\ f'(x) & = \color{#3D99F6}{\frac 1{\ln x} \cdot \frac 1x} \cdot \color{#D61F06}{\frac 1{\ln x} - \frac 1{\ln^2 x} \cdot \frac 1x} \cdot \color{#3D99F6}{\ln (\ln x)} \\ & = \frac 1{x\ln^2 x} - \frac {\ln (\ln x)}{x\ln^2 x} \\ \implies f'(e) & = \frac 1{e\ln^2 e} - \frac {\ln (\ln e)}{e\ln^2 e} \\ & = \frac 1{e(1^2)} - \frac {\ln 1}{e(1^2)} \\ & = \frac 1e = e^{-1} \end{aligned}$

$\implies a = \boxed{-1}$ .