## Calculus: Early Transcendentals 8th Edition

$\frac{d}{dx}(\tan x -x+C)$ $=\sec^2 x-1$ $=\tan^2 x$ (as needed).
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).