Answer
$$f'\left( z \right) = - \frac{1}{{2\sqrt z \left( {1 + z} \right)}}$$
Work Step by Step
$$\eqalign{
& f\left( z \right) = {\cot ^{ - 1}}\sqrt z \cr
& {\text{find the derivative}} \cr
& f'\left( z \right) = \frac{d}{{dz}}\left[ {{{\cot }^{ - 1}}\sqrt z } \right] \cr
& {\text{use }}\frac{d}{{dz}}\left[ {{{\cot }^{ - 1}}u} \right] = - \frac{1}{{1 + {u^2}}}\frac{{du}}{{dz}}.{\text{ let }}u = \sqrt z \cr
& f'\left( z \right) = - \frac{1}{{1 + {{\left( {\sqrt z } \right)}^2}}}\frac{d}{{dz}}\left[ {\sqrt z } \right] \cr
& {\text{then}} \cr
& f'\left( z \right) = - \frac{1}{{1 + {{\left( {\sqrt z } \right)}^2}}}\left( {\frac{1}{{2\sqrt z }}} \right) \cr
& {\text{simplifying}} \cr
& f'\left( z \right) = - \frac{1}{{1 + z}}\left( {\frac{1}{{2\sqrt z }}} \right) \cr
& f'\left( z \right) = - \frac{1}{{2\sqrt z \left( {1 + z} \right)}} \cr} $$
