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 - Page 125: 21


$r'(x) = (4-x)^{-\frac{3}{2}}$ So, $r'(0) = \frac{1}{8}$

Work Step by Step

$r(x) = \frac{2}{\sqrt (4-x)}$ $r'(x) = 2(-\frac{1}{2})(4-x)^{-\frac{3}{2}}(-1)$ $r'(x) = (4-x)^{-\frac{3}{2}}$ So, $r'(0) = \frac{1}{8}$
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