Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt x \left( {1 + x} \right)}}$$
Work Step by Step
$$\eqalign{
& y = {\cot ^{ - 1}}\left( {\sqrt x } \right) \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\cot }^{ - 1}}\left( {\sqrt x } \right)} \right] \cr
& {\text{Apply }}\frac{d}{{dx}}\left[ {{{\cot }^{ - 1}}u} \right] = - \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{1 + {{\left( {\sqrt x } \right)}^2}}}\frac{d}{{dx}}\left[ {\sqrt x } \right] \cr
& {\text{Differentiate and simplify}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{1 + x}}\left( {\frac{1}{{2\sqrt x }}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt x \left( {1 + x} \right)}} \cr} $$
