Answer
\[\frac{-1}{t\sqrt{t^2-1}}\]
Work Step by Step
\[y=R(t)=arc\sin \left(\frac{1}{t}\right)\]
\[\Rightarrow y=\sin^{-1}\left(\frac{1}{t}\right)\;\;\;...(1)\]
Differentiating (1) with respect to $t$ using chain rule
\[\Rightarrow y'=\frac{1}{\sqrt{1-\frac{1}{t^2}}}\cdot \left(\frac{1}{t}\right)'\]
\[\Rightarrow y'=\frac{t}{\sqrt{t^2-1}}\cdot \left(\frac{-1}{t^2}\right)\]
\[\Rightarrow y'=\frac{-1}{t\sqrt{t^2-1}}\]
Hence , \[y'=\frac{-1}{t\sqrt{t^2-1}}\]
