Answer
$\dfrac {1}{\cos ^{2}x}-\dfrac {1}{1+\cos x}
$
Work Step by Step
$y=\dfrac {\tan x}{1+\cos x}\Rightarrow y'=\dfrac {\left( \dfrac {d}{dx}\left( \tan x\right) \right) \times \left( 1+\cos x\right) -\left( \dfrac {d}{dx}\left( 1+\cos x\right) \right) \tan x}{\left( 1+\cos x\right) ^{2}}=\dfrac {\dfrac {1}{\cos ^{2}x}\left( 1+\cos x\right) +\sin x\tan x}{\left( 1+\cos x\right) ^{2}}=\dfrac {1+\cos x+\left( 1+\cos x\right) \left( 1-\cos x\right) \cos x}{\cos ^{2}x\left( 1+\cos x\right) ^{2}}=\dfrac {1}{\cos ^{2}x}-\dfrac {1}{1+\cos x}$
