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: 6

Answer

$r'(s) = \frac{1}{\sqrt {2s+1}}$ $r'(0) = 1$ $r'(1) = \frac{1}{\sqrt 3}$ $r'(\frac{1}{2}) = \frac{1}{\sqrt 2}$

Work Step by Step

$r(s) = \sqrt (2s+1)$ Using the derivative definition: $r'(s) = \lim\limits_{h \to 0}\frac{\sqrt (2(s+h)+1) - \sqrt (2s+1)}{h}$ $r'(s) = \lim\limits_{h \to 0}\frac{(2(s+h)+1) - (2s+1)}{h(\sqrt (2(s+h)+1) + \sqrt (2s+1))}$ $r'(s) = \lim\limits_{h \to 0}\frac{2}{(\sqrt (2(s+h)+1) + \sqrt (2s+1))}$ $r'(s) = \frac{1}{\sqrt (2s+1)}$ Now, $r'(0) = 1$ $r'(1) = \frac{1}{\sqrt 3}$ $r'(\frac{1}{2}) = \frac{1}{\sqrt 2}$
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