How to compute the derivative of $x^x$ using the definition I want to prove that $\displaystyle\lim_{h\to 0}\frac{(x+h)^{x+h}-x^x}{h}=x^x(\ln(x)+1).$
If I write $x^x$ as $e^{x\ln(x)}$ I get: $\displaystyle\lim_{h\to0}\frac{e^{(x+h)\ln(x+h)}-e^{x\ln(x)}}{h}$ but then I'm stuck. What are the next steps? Thanks in advance. 
 A: Here is a simple evaluation of the derivative.
\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}
In the above we have used the substitutions $$t = (x + h)\log(x + h) - x\log x, v = h/x$$ so that both $t, v$ tend to $0$ as $h \to 0$. 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 $a(x)^{b(x)}$.
A: $$\displaystyle\lim_{h\to0}\frac{e^{(x+h)\ln(x+h)}-e^{x\ln(x)}}{h} = \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{(x+h)\ln(x+h)- x \ln(x)}-1}{h}$$
$$ = \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(\frac{x+h}{x})+h \ln(x+h)}-1}{h} \quad = \quad \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(\frac{x+h}{x})} e^{h \ln(x+h)}-1}{h} $$
$$\displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(1+\frac{h}{x})}e^{h \ln(x+h)}-1}{h} =^{taylor} \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x \frac{h}{x}}(x+h)^h-1}{h} $$
$$x^x \displaystyle\lim_{h\to0}\frac{e^h (x+h)^h-1}{h} $$
That's how far I'm able to get without using chain rules or similar. Can anyone help me conclude?
