Find the derivative of the function $(x+\sec x)(x-\tan x)$.

Let $f(x) = (x+\sec x)(x-\tan x)$.
Using the product rule,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = (x+\sec x)\frac{d}{dx}(x-\tan x) + (x-\tan x)\frac{d}{dx}(x+\sec x)$
We know that $\frac{d}{dx}(x) = 1$,$\frac{d}{dx}(\tan x) = \sec^2 x$,and $\frac{d}{dx}(\sec x) = \sec x \tan x$.
Substituting these derivatives:
$f'(x) = (x+\sec x)(1 - \sec^2 x) + (x-\tan x)(1 + \sec x \tan x)$
Expanding the terms:
$f'(x) = (x - x\sec^2 x + \sec x - \sec^3 x) + (x + x\sec x \tan x - \tan x - \sec x \tan^2 x)$
$f'(x) = 2x - x\sec^2 x + \sec x - \sec^3 x + x\sec x \tan x - \tan x - \sec x \tan^2 x$