University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises: 23

Answer

$f' = \frac{1}{\sqrt s(\sqrt s + 1)^{2}}$

Work Step by Step

$f = \frac{\sqrt s - 1}{\sqrt s + 1}$ $f' = \frac{(\sqrt s + 1)(\frac{1}{2}s^{-\frac{1}{2}})-(\sqrt s - 1)(\frac{1}{2}s^{-\frac{1}{2}})}{(\sqrt s + 1)^{2}}$ $f' = \frac{(\frac{1}{2}s^{-\frac{1}{2}})(2)}{(\sqrt s + 1)^{2}}$ $f' = \frac{1}{\sqrt s(\sqrt s + 1)^{2}}$
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