## Calculus (3rd Edition)

$$y'= -3t^{-4}\sec^2t^{-3}.$$
Since $y=\tan t^{-3}$, the derivative $y'$, by using the chain rule, is given by $$y'=\sec^2 t^{-3} (-3t^{-4})=-3t^{-4}\sec^2t^{-3}.$$