In a naive attempt to calculate the derivative of $x^x$

I just find that I can never come up with an explanation restricting myself in 1D. However, if I have one more dimension, I have the following explanation.

If we let $g(u,v)=u^v$, then $f(x) = g(x,x)$. By the chain rule in multivariable calculus, we get $$ f'(x) = g_u\cdot \frac{\partial u}{\partial x}+g_v\cdot \frac{\partial v}{\partial x} $$ where $u(x) = v(x)= x$ and $g_u = vu^{v-1}$ and $g_v=u^v\ln u$. Combining everything together one justifies the "naive" attemp (1).

This is one answer I can come up with explaining the coincidence. It seems that there is no hope to work it out without using multivariable calculus. (I would be glad to be proved wrong and see other answers.)