Answer
$y'=2 \tan x\sec^2 x$
Work Step by Step
Given that $y=2 \tan^2 x-\sec^2x$
Apply derivative rules of differentiation:
$f(x)=p'(x)q(x)+p(x)q'(x)$
$y'=\frac{d}{dx}[2 \tan^2 x-\sec^2x]$
$=4 \tan x\sec^2 x-2\sec x (\sec x \tan x)$
Hence, $y'=2 \tan x\sec^2 x$
