## Calculus 8th Edition

$y'=\dfrac{2-\tan x+x\sec^2 x}{(2 -\tan x)^2}$
Given: $y=\dfrac{x}{2 -\tan x}$ On differentiating w.r.t. $x$ , we get Apply product rule. $y'=\dfrac{d}{dx}(\dfrac{x}{2 -\tan x})\\=\dfrac{(2-\tan x)(1)-x(-\sec^2 x)}{(2 -\tan x)^2}\\=\dfrac{2-\tan x+x\sec^2 x}{(2 -\tan x)^2}$