Answer
$y'=\dfrac{2-\tan x+x\sec^{2}x}{(2-\tan x)^{2}}$
Work Step by Step
$y=\dfrac{x}{2-\tan x}$
Differentiate using the quotient rule:
$y'=\dfrac{(2-\tan x)(x)'-(x)(2-\tan x)'}{(2-\tan x)^{2}}=...$
$...=\dfrac{(2-\tan x)(1)-(x)(-\sec^{2}x)}{(2-\tan x)^{2}}=...$
$...=\dfrac{2-\tan x+x\sec^{2}x}{(2-\tan x)^{2}}$
