#### Answer

$$ y'= -3t^{-4}\sec^2t^{-3}.$$

#### Work Step by Step

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

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Published by
W. H. Freeman

ISBN 10:
1464125260

ISBN 13:
978-1-46412-526-3

$$ y'= -3t^{-4}\sec^2t^{-3}.$$

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