Answer
$f'\left( x \right) = \frac{{1 + \frac{1}{2}\sqrt x }}{{{{\left( {1 + \sqrt x } \right)}^2}}}{\text{ and }}f''\left( x \right) = \frac{{ - 3 - \sqrt x }}{{4\sqrt x {{\left( {1 + \sqrt x } \right)}^3}}}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{x}{{1 + \sqrt x }} \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{{1 + \sqrt x }}} \right] \cr
& {\text{Use the quotient rule}} \cr
& f'\left( x \right) = \frac{{\left( {1 + \sqrt x } \right)\frac{d}{{dx}}\left[ x \right] - x\frac{d}{{dx}}\left[ {1 + \sqrt x } \right]}}{{{{\left( {1 + \sqrt x } \right)}^2}}} \cr
& {\text{Computing derivatives}} \cr
& f'\left( x \right) = \frac{{\left( {1 + \sqrt x } \right)\left( 1 \right) - x\left( {\frac{1}{{2\sqrt x }}} \right)}}{{{{\left( {1 + \sqrt x } \right)}^2}}} \cr
& f'\left( x \right) = \frac{{1 + \sqrt x - \frac{1}{2}\sqrt x }}{{{{\left( {1 + \sqrt x } \right)}^2}}} \cr
& f'\left( x \right) = \frac{{1 + \frac{1}{2}\sqrt x }}{{{{\left( {1 + \sqrt x } \right)}^2}}} \cr
& \cr
& {\text{Find }}f'\left( x \right) \cr
& {\text{ }}f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr
& {\text{ }}f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{1 + \frac{1}{2}\sqrt x }}{{{{\left( {1 + \sqrt x } \right)}^2}}}} \right] \cr
& {\text{ }}f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2 + \sqrt x }}{{2{{\left( {1 + \sqrt x } \right)}^2}}}} \right] \cr
& {\text{ }}f''\left( x \right) = \frac{{2{{\left( {1 + \sqrt x } \right)}^2}\frac{d}{{dx}}\left[ {2 + \sqrt x } \right] - \left( {2 + \sqrt x } \right)\frac{d}{{dx}}\left[ {2{{\left( {1 + \sqrt x } \right)}^2}} \right]}}{{{{\left[ {2{{\left( {1 + \sqrt x } \right)}^2}} \right]}^2}}} \cr
& {\text{Computing derivatives use the chain rule for }}\frac{d}{{dx}}\left[ {2{{\left( {1 + \sqrt x } \right)}^2}} \right] \cr
& {\text{ }}f''\left( x \right) = \frac{{2{{\left( {1 + \sqrt x } \right)}^2}\left( {\frac{1}{{2\sqrt x }}} \right) - \left( {2 + \sqrt x } \right)\left[ {4\left( {1 + \sqrt x } \right)} \right]\left( {\frac{1}{{2\sqrt x }}} \right)}}{{4{{\left( {1 + \sqrt x } \right)}^4}}} \cr
& {\text{Simplifying}} \cr
& {\text{ }}f''\left( x \right) = \frac{{2\left( {1 + \sqrt x } \right)\left[ {1 + \sqrt x - 2\left( {2 + \sqrt x } \right)} \right]}}{{8\sqrt x {{\left( {1 + \sqrt x } \right)}^4}}} \cr
& {\text{ }}f''\left( x \right) = \frac{{1 + \sqrt x - 2\left( {2 + \sqrt x } \right)}}{{4\sqrt x {{\left( {1 + \sqrt x } \right)}^3}}} \cr
& {\text{ }}f''\left( x \right) = \frac{{1 + \sqrt x - 4 - 2\sqrt x }}{{4\sqrt x {{\left( {1 + \sqrt x } \right)}^3}}} \cr
& {\text{ }}f''\left( x \right) = \frac{{ - 3 - \sqrt x }}{{4\sqrt x {{\left( {1 + \sqrt x } \right)}^3}}} \cr
& \cr
& {\text{Summary}} \cr
& f'\left( x \right) = \frac{{1 + \frac{1}{2}\sqrt x }}{{{{\left( {1 + \sqrt x } \right)}^2}}}{\text{ and }}f''\left( x \right) = \frac{{ - 3 - \sqrt x }}{{4\sqrt x {{\left( {1 + \sqrt x } \right)}^3}}} \cr} $$
