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