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: 6

Answer

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

Work Step by Step

$y = \frac{x^{3}}{3} + \frac{x^{2}}{2}+ e^{-x}$ $y' = \frac{3x^{3-1}}{3} + \frac{2x^{2-1}}{2} - e^{-x}$ $y' = x^{2} + x - e^{-x}$ $y'' = 2x +1 +e^{-x}$
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