TITLE: Derivative of $x^x$ at $x=1$ from first principles
QUESTION [6 upvotes]: Find the derivative of $x^x$ at $x=1$ by definition (i.e. using the limit of the incremental ratio).

The only trick I know is $x^x = e^{x \ln x}$ but it doesn't work.

REPLY [13 votes]: Using the definition:
$$
\begin{align}
f'(1)&=\lim_{x\rightarrow1}\frac{x^x-1}{x-1}\\
&=\lim_{x\rightarrow1}\frac{e^{x\log{x}}-1}{x-1}\\
&=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{y}\\
&=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{(1+y)\log(1+y)}\frac{(1+y)\log(1+y)}{y}\\
&=\lim_{t\rightarrow0}\frac{e^{t}-1}{t}\lim_{y\rightarrow0}(1+y)\frac{\log(1+y)}{y}=1
\end{align}
$$
where $y=x-1$, and $t=(1+y)\log(1+y).$