University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 8

Answer

$s' = 2t^{-2}-8t^{-3}$ $s' = -4t^{-3}+ 24t^{-4}$

Work Step by Step

$s= -2t^{-1} + \frac{4}{t^{2}}$ $s' = 2t^{-2}-8t^{-3}$ $s' = -4t^{-3}+ 24t^{-4}$
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