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: 5

Answer

$y ' = 4x^{2} - 1 + 2e^{x}$ $y'' = 8x +2e^{x}$

Work Step by Step

$y = \frac{4x^{3}}{3} -x + 2e^{x}$ $y ' = \frac{4(3)x^{3-1}}{3} - 1 + 2e^{x}$ $y ' = 4x^{2} - 1 + 2e^{x}$ $y'' = 8x^{2-1} - 0 +2e^{x}$ $y'' = 8x +2e^{x}$
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