Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises - Page 267: 54

Answer

$f^{(n)}(x) = \frac{(2\cdot3\cdot4\cdot ...\cdot n)}{(2-x)^{n+1}}=\frac{n!}{(2-x)^{n+1}}$

Work Step by Step

$f(x) = \frac{1}{2-x}$ $f'(x) = \frac{1}{(2-x)^2}$ $f''(x) = \frac{2}{(2-x)^3}$ $f'''(x) = \frac{(2\cdot3)}{(2-x)^4}$ $f^{(4)}(x) = \frac{(2\cdot3\cdot4)}{(2-x)^5}$ etc.... $f^{(n)}(x) = \frac{(2\cdot3\cdot4\cdot ...\cdot n)}{(2-x)^{n+1}}$
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