Answer
$$f'\left( x \right) = \frac{{\sqrt x {e^{\sqrt x }} + 2{e^{\sqrt x }}}}{{2\left( {x{e^{\sqrt x }} + 2} \right)}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \left( {x{e^{\sqrt x }} + 2} \right) \cr
& {\text{differentiate with respect to }}x \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {x{e^{\sqrt x }} + 2} \right)} \right] \cr
& {\text{use }}\frac{d}{{dx}}\ln g\left( x \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}} \cr
& f'\left( x \right) = \frac{{\frac{d}{{dx}}\left[ {x{e^{\sqrt x }} + 2} \right]}}{{x{e^{\sqrt x }} + 2}} \cr
& f'\left( x \right) = \frac{{\frac{d}{{dx}}\left[ {x{e^{\sqrt x }}} \right] + \frac{d}{{dx}}\left[ 2 \right]}}{{x{e^{\sqrt x }} + 2}} \cr
& {\text{use product rule}} \cr
& f'\left( x \right) = \frac{{x\frac{d}{{dx}}\left[ {{e^{\sqrt x }}} \right] + {e^{\sqrt x }}\frac{d}{{dx}}\left[ x \right] + \frac{d}{{dx}}\left[ 2 \right]}}{{x{e^{\sqrt x }} + 2}} \cr
& {\text{compute derivatives }} \cr
& f'\left( x \right) = \frac{{x{e^{\sqrt x }}\left( {\frac{1}{{2\sqrt x }}} \right) + {e^{\sqrt x }}\left( 1 \right) + 0}}{{x{e^{\sqrt x }} + 2}} \cr
& f'\left( x \right) = \frac{{\frac{{\sqrt x {e^{\sqrt x }}}}{2} + {e^{\sqrt x }}}}{{x{e^{\sqrt x }} + 2}} \cr
& f'\left( x \right) = \frac{{\sqrt x {e^{\sqrt x }} + 2{e^{\sqrt x }}}}{{2\left( {x{e^{\sqrt x }} + 2} \right)}} \cr} $$
