## Calculus: Early Transcendentals (2nd Edition)

$\frac{32u^2+8u+1}{(8u+1)^2}$
Quotient Rule states: $(\frac{f}{g})' = \frac{f'g-fg'}{g^2}$ $\frac{d}{du}(\frac{4u^2+u}{8u+1})$ = $\frac{(8u+1)(8u+1)-(4u^2+u)(8)}{(8u+1)^2}$ = $\frac{32u^2+8u+1}{(8u+1)^2}$