I'm working on an optimization problem that takes a huge amount of time to solve if I'm unable to compute the gradient of the objective function.
However, I'm a bit unsure of how to proceed in these gradient calculations - I'm stuck in the calculation of (and in the question if it is possible at all?):
$$ \frac{\partial w^-}{\partial q}(t;q) $$
where,
$$ w^-(t;q) = \inf \left\{W(s;q)\mid s\in [\tau,t] \right\} $$
for some $0\leq \tau < t$ and some differentiable function $W$.
I have tried to apply the chain-rule:
$$ \frac{\partial w^-}{\partial q}(t;q) = \frac{\partial W}{\partial q}(t;q) \cdot \frac{\partial w^-}{\partial W}(t;q) $$
where the gradient $\frac{\partial W}{\partial q}(t;q)$ is computable.
In the objective function I'm integrating this function across a time-horizon $[0,T]$ - hence, the gradient, $\frac{\partial w^-}{\partial q}(t;q)$, doesn't need to be defined for every $t$.
SOLUTION
After trying a couple of things I think I solved the problem:
Let $s^*$ denote the time-point where $W(\cdot;q)$ assumes its minimum in $[\tau,t]$ - then,
$$ \frac{\partial w^-}{\partial q}(t;q) = \frac{\partial W}{\partial q}(s^*;q) $$
which is computable in my problem.
