## Calculus: Early Transcendentals 8th Edition

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