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