Differentiation using the Chain Rule $$y=\frac { \cos(1+x) }{ 1+\cos(x) } $$
Steps I took:
$$y'=\frac { (-\sin(1+x))(1+\cos(x))-(\cos(1+x))(-\sin(x)) }{ (1+\cos(x))^2 } $$
$$y'=\frac { (\cos(1+x))(\sin(x)) }{ (1+\cos(x))^2 } -\frac { (\sin(1+x))(1+\cos(x)) }{ (1+\cos(x))(1+\cos(x)) } $$
$$y'=\frac { (\cos(1+x))(\sin(x)) }{ (1+\cos(x))^2 } -\frac { \sin(1+x) }{1+\cos(x)}$$
How would I further simplify my answer from here? I imagine that I would have to use the trigonometric addition formulas but as soon as I start breaking the problem down to do that, I end up with the incorrect answer. I was wondering if there was a more efficient way of getting to the most simplified solution.

you can do this with without chain rule. let us see. 
$\begin{align}
\dfrac{\cos (1+x)}{1+\cos x} &= \dfrac{\cos 1 \cos x - \sin 1 \sin x}{1+\cos x}\\
&= \dfrac{\cos 1 (1+\cos x) -\cos 1 - \sin 1 \sin x}{1+\cos x}\\
&= \cos 1 - \dfrac{\cos 1}{1 + \cos x} -\sin 1\tan(x/2)
\end{align}$
we can now take the derivative of 
$\begin{align}
\dfrac{d}{dx}\left(\dfrac{\cos (1+x)}{1+\cos x}\right) &=
-\dfrac{\cos 1\sin x}{(1 + \cos x)^2} -\dfrac{\sin 1}{2\cos^2 x/2}\\
&=-\dfrac{\cos 1\sin x}{(1 + \cos x)^2} -\dfrac{\sin 1}{1+\cos x}\\
&=-\dfrac{\cos 1\sin x + \sin 1 + \sin 1\cos x}{(1 + \cos x)^2}\\
&=-\dfrac{\sin (1+x) + \sin 1 }{(1 + \cos x)^2}\\
\end{align}$
i used the calculator to check the derivative and it seems correct. 
