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: 26



Work Step by Step

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