Answer
$\frac{4e^x}{(2e^x+1)^2}$
Work Step by Step
$\left(\frac{f}{g}\right)'$ = $\frac{f'g-fg'}{g^2}$
$\left(\frac{2e^x-1}{2e^x+1}\right)'$ = $\frac{(2e^x)(2e^x+1)-(2e^x-1)(2e^x)}{(2e^x+1)^2}$ = $\frac{4e^x}{(2e^x+1)^2}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
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
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