Answer
$$y'=- \frac{1+\sqrt{1-x^2}}{ \sqrt{1-x^2}\sqrt{1-(x+\sin^{-1}x)^2}}.$$
Work Step by Step
Since $ y=\cos^{-1}(x+\sin^{-1}x) $, then
$$ y'=- \frac{1}{ \sqrt{1-(x+\sin^{-1}x)^2}}(x+\sin^{-1}x)'\\
=- \frac{1+1/\sqrt{1-x^2}}{ \sqrt{1-(x+\sin^{-1}x)^2}}\\
=- \frac{1+\sqrt{1-x^2}}{ \sqrt{1-x^2}\sqrt{1-(x+\sin^{-1}x)^2}}.$$
