Answer
$$ y'= \frac{1-t}{\sqrt{1-t^2}} .$$
Work Step by Step
Since $ y=\sqrt{1-t^2}+\sin^{-1}t $, then the derivative $ y'$ is given by
$$ y'= \frac{(-t^2)'}{2\sqrt{1-t^2}}+\frac{1}{\sqrt{1-t^2}} = \frac{-2t}{2\sqrt{1-t^2}}+\frac{1}{\sqrt{1-t^2}}\\
= \frac{1-t}{\sqrt{1-t^2}} .$$
