University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises - Page 125: 25


$g'(x) = \frac{-1}{(x-1)^{2}}$

Work Step by Step

$g(x) = \frac{x}{x-1}$ $g'(x) = \lim\limits_{z \to x}\frac{\frac{z}{z-1} - \frac{x}{x-1}}{z-x}$ $g'(x) = \lim\limits_{z \to x}\frac{1}{z-x}\frac{x-z}{(x-1)(z-1)}$ $g'(x) = \lim\limits_{z \to x}\frac{-1}{(x-1)(z-1)}$ $g'(x) = \frac{-1}{(x-1)^{2}}$
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