Answer
$y'=\dfrac{\sec^{2}x+\cos x\sec^{2}x+\tan x\sin x}{(1+\cos x)^{2}}$
Work Step by Step
$y=\dfrac{\tan x}{1+\cos x}$
Start the differentiation process by using the quotient rule:
$y'=\dfrac{(1+\cos x)(\tan x)'-(\tan x)(1+\cos x)'}{(1+\cos x)^{2}}=...$
Evaluate the derivatives indicated in the numerator:
$...=\dfrac{(1+\cos x)(\sec^{2}x)-(\tan x)(-\sin x)}{(1+\cos x)^{2}}=...$
$...=\dfrac{\sec^{2}x+\cos x\sec^{2}x+\tan x\sin x}{(1+\cos x)^{2}}$