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: 26

Answer

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

Work Step by Step

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