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 - Page 223: 27


$f'(x) = \frac{2x-1-(x-1)~ln(x-1)}{[1-ln(x-1)]^2~(x-1)}$ The domain is $~~(1, e+1)\cup (e+1, \infty)$

Work Step by Step

$f(x) = \frac{x}{1-ln(x-1)}$ We can differentiate $f(x)$: $f'(x) = \frac{1-ln(x-1)-(x)(-\frac{1}{x-1})}{[1-ln(x-1)]^2}$ $f'(x) = \frac{1-ln(x-1)+\frac{x}{x-1}}{[1-ln(x-1)]^2}$ $f'(x) = \frac{\frac{(1-ln(x-1))(x-1)}{x-1}+\frac{x}{x-1}}{[1-ln(x-1)]^2}$ $f'(x) = \frac{x-x~ln(x-1)-1+ln~(x-1)+x}{[1-ln(x-1)]^2~(x-1)}$ $f'(x) = \frac{2x-1-(x-1)~ln(x-1)}{[1-ln(x-1)]^2~(x-1)}$ When we consider $ln(x-1)$, then it is required that $(x-1)\gt 0$ Then $x \gt 1$ Also: $1-ln(x-1) \neq 0$ $ln(x-1) \neq 1$ $x-1 \neq e$ $x \neq e+1$ The domain is $~~(1, e+1)\cup (e+1, \infty)$
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