Derivative of $(\ln x)^{\ln x}$ How can I differentiate the following function? $$f(x)=(\ln x)^{\ln x}.$$ Is it a composition of functions? And if so, which functions?
Thank you.
 A: The function $f$ can be expressed as
$$
f(x)=(\ln x)^{\ln x}=\mathrm{e}^{(\ln x)\ln(\ln x)}.
$$
Hence $f(x)=\exp\big(g(x)\big)$, where $g(x)=(\ln x)\ln(\ln x)$, and thus
$$
f'(x)=\left(\mathrm{e}^{(\ln x)\ln(\ln x)}\right)'=\mathrm{e}^{(\ln x)\ln(\ln x)}
\left((\ln x)\ln(\ln x)\right)'=(\ln x)^{\ln x}\left(\frac{1}{x}\ln(\ln x)+\ln x\frac{1}{x\ln x}\right).
$$
A: Note that $$a^b = {\left(e^{\ln a}\right)}^b = e^{b\ln a}$$
so $$\left(\ln x\right)^{\ln x} = {\left(e^{\ln\ln x}\right)}^{\ln x}\\
= e^{\ln x\cdot\ln\ln x} = x^{\ln \ln x}$$ either of which which you should be able to do with methods you already know.
We can apply this technique generally to calculate the derivative of $f(x)^{g(x)}$:
$$f(x)^{g(x)} = e^{g\ln f}$$
so $$\begin{align}
\frac d{dx} f^g  = \frac d{dx} e^{g\cdot\ln f} & = \left(\frac d{dx} \left(g\cdot\ln f\right)\right)e^{g\cdot\ln f}\qquad\text{(chain rule)}\\& = \left(\frac d{dx} \left(g\cdot\ln f\right)\right)f^g \\
& = \left(g'\ln f + \frac {gf'}f\right)f^g\qquad\text{(product rule)}
\end{align} $$
where $f' =\frac{df}{dx}$ and $g' = \frac{dg}{dx}$.
A: We define $f(x):= x^x \Rightarrow (f\circ\ln)(x)=\ln(x)^{\ln(x)}$
And by the chain rule $f'(x)=(x^x)'=(e^{x\ln(x)})'= (x\ln(x))'e^{x\ln(x)}=e^{x\ln(x)}(\ln(x)+1)= (\ln(x)+1)x^x$  
Now also by the chain rule $(f\circ\ln)'(x) = f'(\ln(x))\cdot(\ln(x))'=((\ln(\ln(x))+1)\ln(x)^{\ln(x)})\cdot \frac{1}{x} = \frac{(\ln(\ln(x))+1)\ln(x)^{\ln(x)}}{x}$
