Does the generalised directional derivative satisfy any version of the chain rule? 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.
 A: According to theorem 8.14 of https://arxiv.org/pdf/1708.04180.pdf, we have that for locally Lipschitz $g:Y\to \mathbb R$ and Frechet-differentiable $f: X \to Y$, that
$$(g\circ f)^\circ(x;v) \leq g^\circ(f(x);f'(x)v).$$
Equality holds if $g$ is regular.
We can furthermore say that:
$$(g\circ f)^\circ(x;v) \geq -g^\circ(f(x);-f'(x)v)$$
under the same conditions.
A: 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.
A: The straightforward generalization of the usual 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 $g(x)=y$.
