Answer
$\sec^2 (5x-1)$
Work Step by Step
Simplify: $\dfrac{d(\dfrac{1}{5}\tan(5x-1)+C)}{dx}$
Thus, we have
$\dfrac{d(\dfrac{1}{5}\tan(5x-1)+C)}{dx}=(\dfrac{1}{5})\dfrac{5}{\cos^2(5x-1)}=\sec^2 (5x-1)$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 73
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