Answer
\[y'=\frac{-1}{\sqrt{1-(\sin^{-1}t)^2}}\cdot \left(\frac{1}{\sqrt{1-t^2}}\right)\]
Work Step by Step
\[y=\cos^{-1}\left(\sin^{-1}t\right)\;\;\;...(1)\]
We know that \[\frac{d}{dx}(\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}...(2)\] and
\[\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}...(3)\]
Differentiating (1) with respect to $t$ using chain rule
Using (2)
\[y'=\frac{-1}{\sqrt{1-(\sin^{-1}t)^2}}\cdot \left(\sin^{-1}t\right)'\]
Using (3)
\[y'=\frac{-1}{\sqrt{1-(\sin^{-1}t)^2}}\cdot \left(\frac{1}{\sqrt{1-t^2}}\right)\]
Hence \[y'=\frac{-1}{\sqrt{1-(\sin^{-1}t)^2}}\cdot \left(\frac{1}{\sqrt{1-t^2}}\right)\]
