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: 51

Answer

$w' = 6(ze^{2z}+z^{2}e^{2z})=6ze^{2z}(1+z)$ $w'' = 6e^{2z}(1+4z+2z^{2})$

Work Step by Step

$w = 3z^{2}e^{2z}$ $w' = 3(2ze^{2z}+2z^{2}e^{2z})$ $w' = 6(ze^{2z}+z^{2}e^{2z})$ $w'' = 6(e^{2z}+2ze^{2z} + 2ze^{2z}+2z^{2}e^{2z})$ $w'' = 6(e^{2z}+4ze^{2z} +2z^{2}e^{2z})$ $w'' = 6e^{2z}(1+4z+2z^{2})$
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