Aquí hay una evaluación simple del derivado. \begin{align}
f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\
&= \lim_{h \to 0}\frac{(x + h)^{x + h} - x^{x}}{h}\notag\\
&= \lim_{h \to 0}\frac{\exp\{(x + h)\log(x + h)\} - \exp(x\log x)}{h}\notag\\
&= \exp(x\log x)\lim_{h \to 0}\frac{\exp\{(x + h)\log(x + h) - x\log x\} - 1}{h}\notag\\
&= x^{x}\lim_{h \to 0}\frac{\exp\{(x + h)\log(x + h) - x\log x\} - 1}{(x + h)\log(x + h) - x\log x}\cdot\frac{(x + h)\log(x + h) - x\log x}{h}\notag\\
&= x^{x}\lim_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{h \to 0}\frac{(x + h)\log(x + h) - x\log x}{h}\notag\\
&= x^{x}\cdot 1\cdot\lim_{h \to 0}\frac{(x + h)\log(x + h) - x\log(x + h) + x\log(x + h) - x\log x}{h}\notag\\
&= x^{x}\lim_{h \to 0}\dfrac{h\log(x + h) + x\log\left(1 + \dfrac{h}{x}\right)}{h}\notag\\
&= x^{x}\lim_{h \to 0}\left\{\log(x + h) + \dfrac{\log\left(1 + \dfrac{h}{x}\right)}{\dfrac{h}{x}}\right\}\notag\\
&= x^{x}\left\{\log x + \lim_{v \to 0}\frac{\log(1 + v)}{v}\right\}\notag\\
&= x^{x}(\log x + 1)\notag
\end {align} En lo anterior, hemos utilizado las sustituciones$$t = (x + h)\log(x + h) - x\log x, v = h/x$$ so that both $ t, v$ tend to $ 0$ as $ h \ a 0$. Also the following standard limits are used $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ Note that without the use of these standard limits it is not possible get derivatives from first principles for any function of type $ a (x) ^ {b (x)} $.