Answer
$$\frac{{dy}}{{dt}} = - \frac{1}{{2t\sqrt {t - 1} }}$$
Work Step by Step
$$\eqalign{
& y = {\cot ^{ - 1}}\sqrt {t - 1} \cr
& y = {\cot ^{ - 1}}{\left( {t - 1} \right)^{1/2}} \cr
& {\text{find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{{d\left( {{{\cot }^{ - 1}}{{\left( {t - 1} \right)}^{1/2}}} \right)}}{{dt}} \cr
& {\text{we can use the formula }}\cr
& \frac{{d\left( {{{\sin }^{ - 1}}u} \right)}}{{dt}} = - \frac{1}{{1 + {u^2}}}\frac{{du}}{{dt}}.\,\,\,\left( {{\text{see table 7}}{\text{.3}}} \right). \cr
& {\text{here }}u = {\left( {t - 1} \right)^{1/2}};{\text{then}} \cr
& \frac{{dy}}{{dt}} = - \frac{1}{{1 + {{\left( {{{\left( {t - 1} \right)}^{1/2}}} \right)}^2}}}\frac{{d\left( {{{\left( {t - 1} \right)}^{1/2}}} \right)}}{{dt}} \cr
& \frac{{dy}}{{dt}} = - \frac{1}{{1 + t - 1}}\left( {\frac{1}{{2\sqrt {t - 1} }}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dt}} = - \frac{1}{{2t\sqrt {t - 1} }} \cr} $$
