Answer
$$ y' =\frac{-(1+\cos x)\sin(1+x)+\sin x \cos(1+x)}{(1+\cos x)^2}.$$
Work Step by Step
Since $ y=\frac{\cos(1+x)}{1+\cos x}$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and by using the quotient rule: $(u/v)'=\frac{vu'-uv'}{v^2}$ and recalling that $(\cos x)'=-\sin x $, the derivative $ y'$ is given by
$$ y'=\frac{-(1+\cos x)\sin(1+x)-\cos(1+x)(-\sin x)}{(1+\cos x)^2}\\ =\frac{-(1+\cos x)\sin(1+x)+\sin x \cos(1+x)}{(1+\cos x)^2}.$$
