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:
$F'(x)=99(1+x+x^{2})^{98}(1+x+x^{2})'=...$
$...=99(1+x+x^{2})^{98}(1+2x)$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
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 - Page 204: 8
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