Derivative of x^x/Proof

$\frac d{dx} (f(x)^{g(x)}) = \Big(g'(x)\ln(f(x)) + \frac {g(x)f'(x)}{f(x)} \Big)f(x)^{g(x)}$

Proof

$\frac d{dx} (f(x)^{g(x)})$

$= \frac d{dx} \big(e^{\ln(f(x)^{g(x)})}\big)$

$= \frac d{dx} \big(e^{g(x)\ln(f(x))}\big)$

Chain rule and $e^x$ derivative:

$= (g(x)\ln(f(x)))' \big(e^{g(x)\ln(f(x))}\big)$

Product rule:

$= (g'(x)\ln(f(x)) + g(x)(\ln(f(x))') \big(e^{\ln(f(x))}\big)^{g(x)}$

$\ln(x)$ derivative and chain rule:

$= \Big(g'(x)\ln(f(x)) + \frac {g(x)f'(x)}{f(x)}\Big) f(x)^{g(x)}$

$\frac d{dx} x^x$

$= \Big(x' \ln(x) + \frac {x \cdot x'}{x}\Big) x^x$

$= (\ln(x) + 1) \cdot x^x$