More Calculus Than It First Appears

Solution and bonus:

$x^x=(e^{lnx})^{\frac{1}{x}}=e^{\frac{1}{x}lnx}$

$D(e^{\frac{1}{x}lnx})=e^{\frac{1}{x}lnx}\times{(\frac{1}{x}\times{\frac{1}{x}}+{\frac{-1}{x^2}}\times{lnx})}=e^{\frac{1}{x}lnx}\times{(\frac{1}{x^2})(1-lnx)}$

$e^{\frac{1}{x}lnx}$ and $\frac{1}{x^2}$ are always positive, with positive $x$ . When $x<e$ , $1-lnx>0$ , when $x=e$ , $1-lnx=0$ and when $x>e$ , $1-lnx<0$ . Which means that the maximum value is got when $\boxed{x=e}$ , which is the answer for the bonus.

It's obvious that $\sqrt2=\sqrt[4]4$ . Because the maximum value is between $x=2$ and $x=4$ , the biggest of the above must be $\boxed{\sqrt[3]3}$ .

$\sqrt{2} \approx 1.41$

$\sqrt[3]{3} \approx 1.44$

$\sqrt[4]{4} \approx 1.41$

$\sqrt[5]{5} \approx 1.38$

For the bonus, I'll let someone else answer!