Answer
$\frac{d}{dx}(\tan x -x+C)$
$=\sec^2 x-1$
$=\tan^2 x$ (as needed).
Work Step by Step
Differentiating the right side:
$\frac{d}{dx}(\tan x -x+C)$
$=\sec^2 x-1$
Recall the identity $\tan^2 x + 1 = \sec^2x$
$=\tan^2 x$ (as needed).
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 408: 3
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