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: 11

Answer

$r' = \frac{-2}{3s^{3}} + \frac{5}{2s^{2}}$ $r''= 2s^{-4} - 5s^{-3}$

Work Step by Step

$r = \frac{1}{3s^{2}}- \frac{5}{2s}$ $r = \frac{s^{-2}}{3} - \frac{5s^{-1}}{2}$ $r' = \frac{-2s^{-3}}{3} + \frac{5s^{-2}}{2}$ $r' = \frac{-2}{3s^{3}} + \frac{5}{2s^{2}}$ $r''= \frac{6s^{-4}}{3} - \frac{10s^{-3}}{2}$ $r''= 2s^{-4} - 5s^{-3}$
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