Derivative of (ln x)^x

Key idea. The variable sits in both the base and the exponent, so take logarithms before differentiating.

For $x>1$ we have $\ln x>0$, so $f(x)>0$ and we may write

Differentiate with the product and chain rules:

Multiplying through by $f(x)$ gives

At $x=e$ we have $\ln e = 1$, so $(\ln e)^e = 1$, $\ln(\ln e)=\ln 1 = 0$ and $1/\ln e = 1$. Hence

Alternative. Write $f(x)=e^{x\ln\ln x}$. Then $f'(x)=e^{x\ln\ln x}\bigl(\ln\ln x + 1/\ln x\bigr)$, which is the same expression and gives the same value at $x=e$.