Is there a chain rule of any kind for the generalised directional derivative (of the Clarke type)? There is certainly a chain rule for the generalised gradient.

The generalised directional derivative is: $$f^\circ(x;v)=\limsup\_{y \to x, t \downarrow 0} \frac{f(y+tv) - f(y)}{t},$$
where $x,v \in \mathbb R^n$ for some $n$, and $f:\mathbb R^n \to \mathbb R^m$. Albeit, the definition is valid over any Banach space (but I'd like to keep things to finite dimensions for simplicity's sake).

Some information about it can be found here: <https://www.encyclopediaofmath.org/index.php/Clarke_generalized_derivative>

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The naive version of the chain rule is false: Consider $f(x) = |x|, g(x) = -x$. We have that $(f\circ g)^\circ(0;1) = 1$ while $f^\circ(g(0); g^\circ(0;1)) = f^\circ(0;-1) = -1$. What I'm looking for must therefore be an inequality.
['https://mathoverflow.net/questions/355666', 'https://mathoverflow.net', 'https://mathoverflow.net/users/75761/']
A nice and non-trivial extension of the chain-rule occurs from the DiPerna-Lions theory of rough vector fields: take on an open subset $\Omega$ of $\mathbb R^n$ a vector field $X$ with $L^\infty\_{loc}(\Omega)$ coefficients and null divergence such that $X\in W^{1,1}\_{loc}(\Omega)$. Let $u$ be an $L^\infty\_{loc}(\Omega)$ function such that $Xu\in L^1\_{loc}(\Omega)$. Then 
$$
X(u^2)=2u Xu.
$$
The previous chain-rule formula is the main point to prove uniqueness of weak solutions for these vector fields. There are generalizations in several directions: instead of looking at $u^2$, you may check
$$
X(F(u))=F'(u) Xu, \quad F\in C^1.
$$
Also, you can relax the regularity $W^{1,1}$ to $BV$, play a bit with regularity $W^{1,p}$ and relax as well the condition on the divergence by requiring only absolute continuity wrt Lebesgue measure.
The straightforward generalization of the [usual chain rule](https://en.wikipedia.org/wiki/Gradient#Chain_rule) would give 
$$(f\circ g)^\circ(x,v)=v\cdot\bigl(Dg(x)\bigr)^{\rm T}\cdot\bigl(\nabla f(y)\bigr),$$
with $Dg$ the Jacobian matrix and $g(x)=y$.