## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# 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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