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>

[edit]

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/']
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.
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 $g(x)=y$.