Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.1 Exercises: 21

Answer

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

Work Step by Step

$f(x)=\frac{1}{x-1}$ $f'(x)=\lim\limits_{h \to 0}\frac{\frac{1}{(x+h)-1}-\frac{1}{x-1}}{h}$ $f'(x)=\lim\limits_{h \to 0}\frac{(x-1)-((x+h)-1)}{(x-1)((x+h)-1)(h)}$ $f'(x)=\lim\limits_{h \to 0}\frac{x-1-x-h+1}{(x-1)((x+h)-1)(h)}$ $f'(x)=\lim\limits_{h \to 0}\frac{-1}{(x-1)((x+h)-1)}$ $f'(x)=\frac{-1}{(x-1)((x+0)-1)}$ $f'(x)=\frac{-1}{(x-1)^{2}}$
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