Answer
$y'=\frac{-1}{t^2\sqrt (1-1/t^2)}$
Work Step by Step
Use the definition of inverse sine and apply the chain rule to find the derivative:
$y'=\frac{1}{\sqrt (1-(1/t)^2)}\times-1/t^2=\frac{-1}{t^2\sqrt (1-1/t^2)}$
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