Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 16

Answer

$g'(x)=(e^{x^{2}-x})(2x-1)$

Work Step by Step

$g(x)=e^{x^{2}-x}$ Differentiate using the chain rule: $g'(x)=(e^{x^{2}-x})(x^{2}-x)'=(e^{x^{2}-x})(2x-1)$
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