Answer
$y' = 2sec^{2}x\ tanx(2-x) - tan^{2}x$
Work Step by Step
$y = (2-x)tan^{2}x$
$y = (2-x)tanx\ tanx$
$y' = (-tanx +(2-x)sec^{2}x)tanx + (2-x)tanx\ sec^{2}x$
$y' = -tan^{2}x + 2sec^{2}x\ tanx- xsec^{2}x\ tanx + 2tanx\ sec^{2}x - xtanx\ sec^{2}x$
$y' = -tan^{2}x +4sec^{2}x\ tanx-2xsec^{2}x\ tanx$
$y' = 2sec^{2}x\ tanx(2-x) - tan^{2}x$
