#### Answer

$\frac{4w}{(w^2+1)^2}$

#### Work Step by Step

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

Published by
Pearson

ISBN 10:
0321947347

ISBN 13:
978-0-32194-734-5

$\frac{4w}{(w^2+1)^2}$

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