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: 22



Work Step by Step

$\left(\frac{f}{g}\right)'$ = $\frac{f'g-fg'}{g^2}$ $\left(\frac{2e^x-1}{2e^x+1}\right)'$ = $\frac{(2e^x)(2e^x+1)-(2e^x-1)(2e^x)}{(2e^x+1)^2}$ = $\frac{4e^x}{(2e^x+1)^2}$
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