Answer
$99!$
Work Step by Step
$P(x)$ = $x^{99}+...$
$P'(x)$ = $99x^{98}+...$
$P''(x)$ = $99\times98x^{97}+...$
$P'''(x)$ = $99\times98\times97x^{96}+...$
$P^{99}(x)$ =$99!x^{0}$ = $99!$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - 3.5 Higher Derivatives - Exercises - Page 137: 53
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