Answer
$\frac{-1}{\sqrt( {2t}-{t^2})}$
Work Step by Step
Given that $y$ = $sin^{-1}(1-{t})$
on substituting ${1-t}$ = $u$
on applying chain rule derivative
$\frac{dy}{dt}$ = $\frac{{dsin^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{\sqrt (1-{u^2})}$ $\times$ $\frac{du}{dt}$
so $\frac{dy}{dt}$ = $\frac{1}{\sqrt (1-{({1-t})^2})}$ $\times$ $\frac{d {(1-t)}}{dt}$
or $\frac{dy}{dt}$ = $\frac{1}{\sqrt( {2t}-{t^2})}$ $\times$ $-1$
The final answer is: $\frac{-1}{\sqrt( {2t}-{t^2})}$