Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises - Page 160: 28


$f'(x) = \frac{-4\sqrt x-4+ \frac{1}{\sqrt x}}{(4x+1)^2}$

Work Step by Step

$f(x) = (2\sqrt x-1)(4x+1)^{-1} = \frac{2\sqrt x - 1}{4x+1}$ $f'(x) = \frac{(\frac{1}{\sqrt x})(4x+1)-(2\sqrt x-1)(4)}{(4x+1)^2} = \frac{4\sqrt x + \frac{1}{\sqrt x} -8\sqrt x-4}{(4x+1)^2} = \frac{-4\sqrt x-4+ \frac{1}{\sqrt x}}{(4x+1)^2}$
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