#### Answer

The Derivative is:
$y'=-\tan^2 x+4\tan x\sec^2 x-2x\tan x\sec^2 x$

#### Work Step by Step

$y=(2-x)\tan^2 x$
$y=(2-x)\tan x\tan x$
Applying Derivative rules:
$y'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$
$y'=(0-x^{1-1})\tan x\tan x+(2-x)((sec^2 x)\tan x+\tan x(\sec^2x))$
$y'=(-1)\tan^2 x+(2-x)(2\tan x\sec^2 x)$
$y'=-\tan^2 x+4\tan x\sec^2 x-2x\tan x\sec^2 x$