Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 53

Answer

$f^{n}(x)= \frac{(-1)^{n-1}(n-1)!}{(x-1)^{n}}$

Work Step by Step

Differentiate $f(x)=ln(x-1)$ with respect to $x$. $f'(x)=\frac{1}{(x-1)}$ $f''(x)=(-1)(x-1)^{-2}$ $f'''(x)=(-1)^{2}1.2(x-1)^{-3}$ After solving in the same manner, we find that $f^{n}(x)=(-1)^{n-1}1.2.3....(n-1)(x-1)^{-n}$ Hence,$f^{n}(x)= \frac{(-1)^{n-1}(n-1)!}{(x-1)^{n}}$
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