$x= \cos ^{-1} \bigg(\large\frac{1}{\sqrt {1+t^2}}\bigg),$$y= \sin ^{-1} \bigg(\large\frac{t}{\sqrt {1+t^2}}\bigg)=>\frac{dy}{dx}=$

Question

\[\begin {array} {1 1} (1)\;0 & \quad (2)\;\tan t \\ (3)\;1 & \quad (4)\;\sin t \cos t \end {array}\]