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: 42

Answer

$y' = \frac{x^{4}}{24}$ $y'' = \frac{x^{3}}{6}$ $y''' = \frac{x^{2}}{2}$ $y'''' = x$ $y''''' = 1$ $y'''''' = 0$ (and all higher derivatives will also be 0)

Work Step by Step

$y = \frac{x^{5}}{120}$ $y' = \frac{x^{4}}{24}$ $y'' = \frac{x^{3}}{6}$ $y''' = \frac{x^{2}}{2}$ $y'''' = x$ $y''''' = 1$ $y'''''' = 0$ (and all higher derivatives will also be 0)
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