## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises: 3

#### 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).

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