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: 25

Answer

$v' = \frac{2\sqrt x-1}{x^{2}}$

Work Step by Step

$v = \frac{1+x-4\sqrt x}{x}$ Quotient rule: $v' = \frac{x(1-2x^{-\frac{1}{2}}) - (1+x-4\sqrt x)}{x^{2}}$ $v' = \frac{x-2x^{\frac{1}{2}} - 1-x+4\sqrt x}{x^{2}}$ $v' = \frac{2\sqrt x-1}{x^{2}}$
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