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: 52

Answer

$w' = 2e^{z}z^{2} - e^{z}z + e^{z}z^{3}=e^z(2z^2-z+z^3)$ $w'' = 5z^{2}e^{z} + 3e^{z}z - e^{z} + e^{z}z^{3}=e^z(5z^2+3z-1+z^3)$

Work Step by Step

$w = e^{z}(z-1)(z^{2}+1)$ $w = e^{z}(z^{3}-z^{2}+z-1)$ $w' = e^{z}(z^{3}-z^{2}+z-1) + e^{z}(3z^{2}-2z+1)$ $w' = 2e^{z}z^{2} - e^{z}z + e^{z}z^{3}$ $w''= 2e^{z}z^{2} +4e^{z}z - e^{z}z-e^{z}+e^{z}z^{3}+3z^{2}e^{z}$ $w'' = 5z^{2}e^{z} + 3e^{z}z - e^{z} + e^{z}z^{3}$
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