Given a function $q=q(t, e)$, where $ e= e(t, v(\tau))$, such that $q = q(t, e(t, v(\tau)))$, how to find the expression for $\frac{dQ}{d \tau}$ if
$$Q(\tau) = \int_0^\tau q(t, e(t, v(\tau))) dt$$
such that
$$ \frac{dQ}{d \tau} = \frac{d}{d\tau} \left( \int_0^\tau q(t, e(t, v(\tau))) dt \right) $$
where the integrand is a function of the integral limits through composition?
I'm thinking that the Leibniz integral rule might come in handy (link below), but I am having troubles applying it due to the composition. Any ideas, tips, or thoughts are appreciated.
Leibniz integral rule link: https://en.wikipedia.org/wiki/Leibniz_integral_rule
Let $$ F(a, b) = \int_0^a q(t, e(t, v(b))) dt $$ Then $Q(\tau) = F(\tau, \tau)$, and using the chain rule from multivariable calculus we have $$ \frac{dQ}{d\tau} = \frac{\partial F}{\partial a}\frac{da}{d\tau} + \frac{\partial F}{\partial b}\frac{db}{d\tau}. $$ Here, $a(\tau) = b(\tau) = \tau$ and hence $$ \frac{dQ}{d\tau} = \frac{\partial F}{\partial a}(\tau, \tau) + \frac{\partial F}{\partial b}(\tau, \tau) \\ = q(t, e(v(\tau))) + \int_0^\tau \frac{\partial q}{\partial e}\frac{\partial e}{\partial v}\frac{dv}{d\tau}dt, $$ where I've assumed all the functions are sufficiently differentiable to allow differentiating under the integral sign.