Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises - Page 268: 67


$\frac{f'(x)}{f(x)} = \frac{1}{(x-a)}+\frac{1}{(x-b)}+\frac{1}{(x-c)}$

Work Step by Step

$f(x) = (x-a)(x-b)(x-c)$ Let $g(x) = (x-a)(x-b)$ and let $h(x) = (x-c)$ Then $f'(x) = g'(x)~h(x)+g(x)~h'(x)$ We can find $f'(x)$: $f'(x) = [(x-b)+(x-a)]\cdot (x-c)+(x-a)(x-b)$ $f'(x) = (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)$ Then: $\frac{f'(x)}{f(x)} = \frac{(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)}{(x-a)(x-b)(x-c)}$ $\frac{f'(x)}{f(x)} = \frac{(x-b)(x-c)}{(x-a)(x-b)(x-c)}+\frac{(x-a)(x-c)}{(x-a)(x-b)(x-c)}+\frac{(x-a)(x-b)}{(x-a)(x-b)(x-c)}$ $\frac{f'(x)}{f(x)} = \frac{1}{(x-a)}+\frac{1}{(x-b)}+\frac{1}{(x-c)}$
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