Let there be a subadditive function $f : \mathbb R^n \to \mathbb R$, that is ,\begin{gather}f(a+b) \leq f(a) + f(b) \quad \forall a, b \in \mathbb R^n,\tag{1}\label{eq1}\end{gather} and lets write $f' : \mathbb R^n \to \mathbb R^n$ for the gradient of $f$.
Question 1: Under which circumstances can we expect the subadditivity property also hold for $f'$, where "$\leq$" is understood component-wise, that is, \begin{gather}f_i'(a+b) \leq f_i'(a) + f_i'(b) \enspace \quad \forall i \in [n],\enspace a, b \in \mathbb R^n\text? \tag{2}\label{eq2}\end{gather}
A popular example for subadditive functions is the square root $f(t) = \sqrt{t}$, which is only defined on a subset of $\mathbb R$, but still preserves the subadditivity property under differentiation.
In my specific case, the function $f$ is additionally convex, hence, we know that $f'_i$ possesses no extrema. It is also defined on the whole $\mathbb R^n$. Does this have implications for Eq. \eqref{eq2}?
Question 2: Out of interest, can somebody provide a counterexample of a function $f$, which is subadditive on the whole $\mathbb R$ (ie, Eq. \eqref{eq1} is fulfilled), but contradicts Eq. \eqref{eq2}?
Consider $f(x) = |x|$ which is subadditive on $\mathbb R$. Its derivative is not defined at $0$, but even where it is defined it fails to be subadditive, e.g. $f'(1) > f'(-1) + f'(2)$.
For an example that is differentiable on all of $\mathbb R$, try $$ f(x) = \cases{2|x| & if $|x|\ge 1$\cr x^2+1 & otherwise}$$
EDIT: Somewhat more generally, the only cases where an even differentiable function $f$ and its derivative $f'$ are both subadditive on all of $\mathbb R$ are when $f$ is constant. Indeed, suppose $f$ is subadditive, differentiable and even, so $f'$ is odd. But then if $f'$ is subadditive it must be additive (and thus linear), since if $f'(x + y) < f'(x) + f'(y)$ we would have $f'(-x-y) = -f'(x+y) > -f'(x) - f'(y) = f'(-x) + f'(-y)$. And since a quadratic is never subadditive on $\mathbb R$, this can only happen if $f$ is constant.