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: 20

Answer

$f'(x) = x^{-2}$ So, $f'(\sqrt 3) = \frac{1}{(\sqrt 3)^{2}} = \frac{1}{3}$

Work Step by Step

$f(x) = 1 - \frac{1}{x}$ Therefore, $f'(x) = 0 - (-1)x^(-1-1)$ $f'(x) = x^{-2}$ So, $f'(\sqrt 3) = \frac{1}{(\sqrt 3)^{2}} = \frac{1}{3}$
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