Derivative of a quotient

We have $u'=-1$ and $v'=1$, so
$$f'(x)=\frac{(-1)(4+x)-(1-x)(1)}{(4+x)^2}=\frac{-4-x-1+x}{(4+x)^2}=\frac{-5}{(4+x)^2}.$$
The function $f$ is defined on $\mathbb{R}\setminus\{-4\}$, and $f'$ is defined on the same set.

Remark. Since $f'<0$ everywhere on its domain, $f$ is decreasing on each of the intervals $(-\infty,-4)$ and $(-4,+\infty)$. Be careful: it is *not* decreasing on the union, as $f(-5)=-6$ is smaller than $f(0)=\frac14$.