Derivatives of trigonometric functions: $y= \frac{x \sin(x)}{1+\cos(x)}$ 
I'm trying to find the derivative of:
$$y= \frac{x \sin(x)}{1+\cos(x)}$$

I've tried but I can't achieve the simplified form -
Here's my try-
$$y' =  \left(\frac{x \sin(x)}{1+\cos(x)}\right)'$$
$$y' = \frac{x\sin^2(x) + (\cos(x)+1 )(\sin(x)+x\cos(x))}{(\cos(x)+1)^2}$$
I'm pretty sure the above is correct that is why I didn't show the steps in between ... but I can't simplify it until -
$$\frac{x+\sin(x)}{1+\cos(x)}$$
Which concept or formula am I missing out from in order to simplify it further? Or what should I do next? Thanks!

Note that $x\sin^2 x = x(1 - \cos^2 x)$. So we can rewrite the numerator as $$x-x\cos^2 x + x\cos^2 x+(1+\cos x)\sin x +x\cos x = (1+\cos x)(x+\sin x)$$ so $$y'=\frac{(1+\cos  x)(x+\sin x)}{(1+\cos x)(1+\cos x)} = \frac{x+\sin x}{1+\cos x}$$ 
