Answer
$f'(x) = \frac{1}{2x~\sqrt{2+ln~x}}$
The domain is $~~(\frac{1}{e^2}, \infty)$
Work Step by Step
$f(x) = \sqrt{2+ln~x}$
We can differentiate $f(x)$:
$f'(x) = \frac{1}{2}(2+ln~x)^{-1/2}\cdot \frac{d}{dx}(2+ln~x)$
$f'(x) = \frac{1}{2~\sqrt{2+ln~x}}\cdot \frac{1}{x}$
$f'(x) = \frac{1}{2x~\sqrt{2+ln~x}}$
When we consider $~~\sqrt{2+ln~x}~~$ it is required that $~~2+ln~x \gt 0$
Then $~~ln~x \gt -2$
Then $~~x \gt e^{-2}$
Then $~~x \gt \frac{1}{e^2}$
The domain is $~~(\frac{1}{e^2}, \infty)$
