Answer
$\frac{4w}{(w^2+1)^2}$
Work Step by Step
$\left(\frac{f}{g}\right)'$ = $\frac{f'g-fg'}{g^2}$
$\frac{w^2-1}{w^2+1}$ = $\frac{(2w)(w^2+1)-(w^2-1)(2w)}{(w^2+1)^2}$ = $\frac{4w}{(w^2+1)^2}$
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: 26
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