## Thomas' Calculus 13th Edition

The Derivative is: $y'=0$
$y=(\sec x+\tan x)(\sec x-\tan x)$ Applying the notable product $(a+b)(a-b)=a^2-b^2$ $y=\sec^2 x-\tan^2 x$ Applying Derivative rules: $y'=\frac{d}{dx}(\sec x\sec x)-\frac{d}{dx}(\tan x\tan x)$ Apply the product rule: $y'=((\sec x\tan x)(\sec x)+(\sec x)(\sec x\tan x))-((\sec^2 x)(\tan x)+(\tan x)(\sec^2 x))$ $y'=2\sec^2 x\tan x - 2\sec^2 x\tan x$ $y'=0$