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: 8

Answer

$F'(x)=99(1+x+x^{2})^{98}(1+2x)$

Work Step by Step

$F(x)=(1+x+x^{2})^{99}$ Differentiate using the chain rule: $F'(x)=99(1+x+x^{2})^{98}(1+x+x^{2})'=...$ $...=99(1+x+x^{2})^{98}(1+2x)$
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