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
$$ y'=\sec^2 t^{-3} (-3t^{-4})=-3t^{-4}\sec^2t^{-3}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 43
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