Derivative of double integral with direct and inverse functions I found a couple of similar question, but I am struggling applying their logic to my example.
Derivative of double integral with respect to upper limits
Differentiation under the double integral sign
Derivative of double integral with respect to upper limits
Derivative of double integral using Leibniz integral rule
I have the following differentiation of the double integral:
$$\frac{d}{dc}\int_{z^{-1}(c)}^{1}\int_{z(x)}^{z^{-1}(c)} (v-y)f(x)f(y)dydx$$
where $f(x)$ and $f(y)$ and probability density functions.
Is it possible to apply Leibniz rule right away here to somehow simplify it? Substituting the internal integral for anti-derivatives as in the examples does not seem possible because that's a product of functions.
The best I can do is to "open up" the internal integral by integrating by parts and then apply the rule:
$$\int_{z(x)}^{z^{-1}(c)} (v-y)f(y)dy=(v-z^{-1}(c))F(z^{-1}(c))-(v-z(x))F(z(x))+\int_{z(x)}^{z^{-1}(c)}F(y)dy$$
Then my expression becomes something like:
$$\frac{d}{dc}\int_{z^{-1}(c)}^{1}(v-z^{-1}(c))F(z^{-1}(c))f(x)dx-\frac{d}{dc}\int_{z^{-1}(c)}^{1}(v-z(x))F(z(x))f(x)dx+$$
$$\frac{d}{dc}\int_{z^{-1}(c)}^{1}\int_{z(x)}^{z^{-1}(c)}F(y)f(x)dydx$$
where the first term seems to be feasible to take, and even the second one, but the third one - back to square one. Is there any way around it?
 A: Let's look at the Leibnitz rule: If $g(r, t)$ and $g_r(r,t)$ are continuous, and $a(r), b(r)$ have continuous derivatives, then
$$\frac{d}{dr}\left(\int_{a(r)}^{b(r)} g(r, t)\,dt\right) = g(r, b(r))\frac{db}{dr} - g(r,a(r))\frac{da}{dr} + \int_{a(r)}^{b(r)}\frac{\partial}{\partial r}g(r, t)\,dt$$
For your problem, $r = c, t= x, b = 1, a(c) = z^{-1}(c)$ and $$g(c, x) = f(x)\int_{z(x)}^{z^{-1}(c)}(v-y)f(y)\, dy$$
Note that no condition of the Leibnitz rule says anything about products of functions. All that it requires is the $g_c(c,x)$ exist and be continuous. $b$ obviously meets its condition. Assuming that $z^{-1}$ is continuously differentiable and that $f$ and $z$ are at least continuous, that expression will satisfy the conditions. Since $b$ is constant, $\frac{db}{dr}$ term is $0$.
Therefore $$\frac{d}{dc}\int_{z^{-1}(c)}^{1}\int_{z(x)}^{z^{-1}(c)} (v-y)f(x)f(y)dydx\\=-\frac{dz^{-1}}{dc}f(z^{-1}(c))\int_{c}^{z^{-1}(c)}(v-y)f(y)\, dy + (v-z^{-1}(c))f(z^{-1}(c))\frac{dz^{-1}}{dc}\int_{z^{-1}(c)}^{1}f(x)\,dx$$
