Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.2 - The Derivative as a Function - Exercises 3.2 - Page 115: 26

Answer

$ \dfrac{1}{2\sqrt x}$

Work Step by Step

$g(x) = 1+\sqrt x$ We will apply definition of derivative, such as: $g'(x) = \lim\limits_{z \to x}\dfrac{1 + \sqrt z - 1-\sqrt x}{(z-x)}$ Then, we have $g'(x) = \lim\limits_{z \to x}\dfrac{ \sqrt z -\sqrt x}{(z-x)}= \lim\limits_{z \to x}\dfrac{1}{\sqrt z+\sqrt x}=\dfrac{1}{\sqrt x+\sqrt x}$ Hence, $g'(x) = \dfrac{1}{2\sqrt x}$
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