University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises - Page 124: 10


$v'(t) = 1+t^{-2}$

Work Step by Step

$v(t) = t - \frac{1}{t}$ Therefore, $v'(t) = (1)t^{1-1}-(-1)t^{-1-1}$ $v'(t) = 1+t^{-2}$
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