Answer
$\frac{32u^2+8u+1}{(8u+1)^2}$
Work Step by Step
Quotient Rule states: $(\frac{f}{g})' = \frac{f'g-fg'}{g^2}$
$\frac{d}{du}(\frac{4u^2+u}{8u+1})$ = $\frac{(8u+1)(8u+1)-(4u^2+u)(8)}{(8u+1)^2}$ = $\frac{32u^2+8u+1}{(8u+1)^2}$
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 - Review Exercises - Page 233: 21
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