Answer
$$f'\left( x \right) = \frac{1}{{2\left( {1 + x} \right)\sqrt x }}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \arctan \sqrt x \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\arctan \sqrt x } \right] \cr
& {\text{Recall that }}\frac{d}{{dx}}\left[ {\arctan u} \right] = \frac{1}{{1 + {u^2}}}\left( {\frac{{du}}{{dx}}} \right){\text{ }}\left( {{\text{Page 371}}} \right) \cr
& f'\left( x \right) = \frac{1}{{1 + {{\left( {\sqrt x } \right)}^2}}}\frac{d}{{dx}}\left[ {\sqrt x } \right] \cr
& f'\left( x \right) = \frac{1}{{1 + x}}\left( {\frac{1}{{2\sqrt x }}} \right) \cr
& f'\left( x \right) = \frac{1}{{2\left( {1 + x} \right)\sqrt x }} \cr} $$
