Answer
$\frac{{dy}}{{dt}} = - \frac{1}{{\sqrt {1 - {t^2}} \sqrt {1 - {{\left( {{{\sin }^{ - 1}}t} \right)}^2}} }}$
Work Step by Step
$$\eqalign{
& y = {\cos ^{ - 1}}\left( {{{\sin }^{ - 1}}t} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{{\cos }^{ - 1}}\left( {{{\sin }^{ - 1}}t} \right)} \right] \cr
& {\text{Use }}\frac{d}{{dt}}\left[ {{{\cos }^{ - 1}}u} \right] = - \frac{1}{{\sqrt {1 - {t^2}} }}\frac{{du}}{{dt}},{\text{ let }}u = {\sin ^{ - 1}}t \cr
& \frac{{dy}}{{dt}} = - \frac{1}{{\sqrt {1 - {{\left( {{{\sin }^{ - 1}}t} \right)}^2}} }}\frac{d}{{dt}}\left[ {{{\sin }^{ - 1}}t} \right] \cr
& {\text{Computing the derivative}} \cr
& \frac{{dy}}{{dt}} = - \frac{1}{{\sqrt {1 - {{\left( {{{\sin }^{ - 1}}t} \right)}^2}} }}\left( {\frac{1}{{\sqrt {1 - {t^2}} }}} \right) \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dt}} = - \frac{1}{{\sqrt {1 - {t^2}} \sqrt {1 - {{\left( {{{\sin }^{ - 1}}t} \right)}^2}} }} \cr} $$
