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 - Page 136: 26


$r' = \frac{1}{\sqrt x}(1-\frac{1}{x})$

Work Step by Step

$r = 2(\frac{1}{\sqrt x} +\sqrt x)$ ($\theta$ is replaced with x for convinience.) $r = 2(x^{-\frac{1}{2}}+ x^{\frac{1}{2}})$ $r' = 2(-\frac{1}{2}x^{-\frac{3}{2}} + \frac{1}{2}x^{-\frac{1}{2}})$ $r' = -x^{-\frac{3}{2}} + x^{-\frac{1}{2}}$ $r' = x^{-\frac{1}{2}}(1-x^{-1})$ $r' = \frac{1}{\sqrt x}(1-\frac{1}{x})$
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