If $f(x) = x^{\sqrt{x}}$ for $x > 0$, what is $f'\left(4\right)$?

Explanation

1. 01

   $\log y = \sqrt{x}\log x$, so $\frac{y'}{y} = \frac{\log x}{2\sqrt{x}} + \frac{\sqrt{x}}{x}$.

2. 02

   At 4: $y = 4^{2} = 16$ and $\frac{y'}{y} = \frac{2\log 2}{4} + \frac{1}{2} = \frac{1 + \log 2}{2}$, so $f'(4) = 8(1 + \log 2)$.