Answer
$y ' = 4x^{2} - 1 + 2e^{x}$
$y'' = 8x +2e^{x}$
Work Step by Step
$y = \frac{4x^{3}}{3} -x + 2e^{x}$
$y ' = \frac{4(3)x^{3-1}}{3} - 1 + 2e^{x}$
$y ' = 4x^{2} - 1 + 2e^{x}$
$y'' = 8x^{2-1} - 0 +2e^{x}$
$y'' = 8x +2e^{x}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 3 - Section 3.3 - Differentiation Rules - Exercises - Page 135: 5
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