Answer
$y' = \frac{1}{{2x\sqrt {x - 1} }}$
Work Step by Step
$$\eqalign{
& y = {\tan ^{ - 1}}\sqrt {x - 1} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\sqrt {x - 1} } \right] \cr
& {\text{Use the Derivatives of Inverse Trigonometric Functions }}\left( {p223} \right) \cr
& {\text{and the chain rule}}{\text{. }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{{u^2} + 1}}\frac{{du}}{{dx}},\;{\text{ }}u = \sqrt {x - 1} \cr
& y' = \frac{1}{{1 + {{\left( {\sqrt {x - 1} } \right)}^2}}}\frac{d}{{dx}}\left[ {\sqrt {x - 1} } \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& y' = \frac{1}{{1 + \left( {x - 1} \right)}}\left( {\frac{1}{{2\sqrt {x - 1} }}} \right) \cr
& y' = \frac{1}{{2x\sqrt {x - 1} }} \cr} $$
