I'm trying to find the derivative to: $$\frac{d}{dt}\int^{a(t)}_{b(t)}\int^{c(x,t)}_{d(x,t)}f(x,y)dydx$$ I don't know how to approach this. I was searching for anything similar on StackExchange and this was the closest question I could find: Derivative of double integral with respect to upper limits . I am hesitant to use this because my bounds with respect to y are functions of x and t (as opposed to just t).
For context, I am trying to solve for the time derivative of magnetic flux in a circular loop to get emf/voltage ($\frac{d}{dt}\iint B(\sqrt{x^2+(h-y)^2})dxdy$). The field is centered at $(0,h)$. The exact function is: $$\varepsilon = \frac{d}{dt}\int^{a(t)+R}_{a(t)-R}\int^{\sqrt{R^2-(x-a(t))^2}+b(t)}_{-\sqrt{R^2-(x-a(t))^2}+b(t)}\frac{1}{1+ce^{-d\sqrt{x^2+(h-y)^2}}}dydx$$ $$a(t)=-hcos(\omega t); b(t)=hsin(\omega t)$$ $a(t)$ and $b(t)$ are the coordinates of the center of the circle. The setup is a disk magnet centered at $(0,h)$ whose B field (only the outward/normal component to its surface) is roughly defined by $B(r)$, where r is the distance between a point and the center of the magnet. A coil of wire is revolving around the origin with angular velocity $\omega$. I have excluded N, the number of loops, and another constant.
I ideally want the derivative as a function. Though, at the very least, I need the extrema.
We could derive a complicated looking expression for the derivative, but that would hardly serve your needs for this problem. Why would you want to do this integral in Cartesian coordinates in the first place?
Go to "polar" coordinates centered at $(a(t),b(t))$:
$$x = r\cos\theta + a(t)$$
$$y = r\sin\theta + b(t)$$
which makes the integral
$$\Phi(t) = \int_0^{2\pi} \int_0^R \frac{r}{1+ce^{-d\sqrt{r^2-2rh(\cos\omega t\cos\theta-\sin\omega t\sin\theta)+h^2}}}\:dr\:d\theta$$
$$= \int_0^{2\pi} \int_0^R \frac{r}{1+ce^{-d\sqrt{r^2-2rh\cos(\theta+\omega t)+h^2}}}\:dr\:d\theta$$
$$= \int_{\omega t}^{2\pi+ \omega t} \int_0^R \frac{r}{1+ce^{-d\sqrt{r^2-2rh\cos\theta+h^2}}}\:dr\:d\theta$$
then use the fundamental theorem of calculus to get your emf:
$$\epsilon = \int_0^R \frac{\omega r}{1+ce^{-d\sqrt{r^2-2rh\cos(2\pi+\omega t)+h^2}}}\:dr - \int_0^R \frac{\omega r}{1+ce^{-d\sqrt{r^2-2rh\cos(\omega t)+h^2}}}\:dr = 0 $$
which shouldn't surprise you. Intuitively from looking at the problem, if I have a loop orbiting in an azimuthally symmetric magnetic field, the flux won't change $-$ it would have only changed if the loop had some radially inward/outward motion. This reasoning was probably the intended solution for a physics class, not the integral way.