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: 24


$u'= \frac{5x-1}{4x\sqrt{x}}$ $u'= \frac{5x-1}{4x^{3/2}}$

Work Step by Step

$u = \frac{5x+1}{2\sqrt x}$ quotient rule: $u' = \frac{2\sqrt x(5) - (5x+1)x^{-\frac{1}{2}}}{4x}$ $u' = \frac{10\sqrt x - 5\sqrt x-x^{-\frac{1}{2}})}{4x}$ $u' = \frac{5\sqrt x-x^{-\frac{1}{2}}}{4x}$ $u'= \frac{5x-1}{4x\sqrt{x}}$ $u'= \frac{5x-1}{4x^{3/2}}$
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