How to calculate $\frac d{dx}x^x$ without differentiating any logarithms? How to calculate $\frac d{dx}x^x$ without differentiating any logarithms?  There isn't any clear derivative rule I can use and the derivative quotients aren't any help:
$$\lim_{h\to0}\frac{f(x+h)-f(x)}h\text{ or }\lim_{h\to x}\frac{f(x)-f(h)}{x-h}$$
I was wondering if there were any ways to do this using either power rule or derivatives of exponential functions.

To be more concise as to what I mean here, as pointed out by the comments, I could just write $x^x=e^{x\ln(x)}$.  However, that would involve taking the derivative of a logarithm.  One may use $\frac d{dx}a^x=\ln(a)a^x$, which seems intuitive, since if we could somehow change the $a$ to an $x$, we'd almost have the correct derivative.  Though I haven't the slightest of how one would formally do that.

First note that we can write
$$(x+h)^{x+h}=x^{x+h}\left(1+\frac hx\right)^{x+h} \tag 1$$
Expanding the parenthetical term on the right-hand side of $(1)$ in a generalized Binomial Series reveals
$$\left(1+\frac hx\right)^{x+h}=1+\frac{h(x+h)}{x}+O(h^2) \tag 2$$
Putting together $(1)$ and $(2)$ yields
$$\frac{(x+h)^{x+h}-x^x}{h}=x^x\left(\frac{x^h-1}{h}\right)+x^xx^h+O(h)$$
Taking the limit as $h\to 0$, we see that
$$\begin{align}
\lim_{h\to 0}\left(\frac{(x+h)^{x+h}-x^x}{h}\right)&=x^x\left(1+\lim_{h\to 0}\left(\frac{x^h-1}{h}\right)\right)\\\\
&=x^x(1+\log(x))
\end{align}$$
And we are done!
