Using Leibniz's rule for second derivatives of Double integrals I have a question on the evaluation of:
$\frac{d^2}{dx^2}\int _0^x\:\int _0^x\:f\left(s,t\right)dsdt$
I am confused on the application of Leibniz's rule here. 
It should be relatively straightforward because it is similar to the FTC, but the correct answer has eluded me.
Basically I changed the variables so $0$ because some function $f(u)$ and then integrated the integrand with respect to $s$, yielding
$\frac{d}{dx}\int _u^x\:\left[f\left(x,t\right)\frac{dx}{dx}-f\left(u,t\right)\frac{du}{dx}\right]dt\:+\:\frac{\partial }{\partial x}\left(\int _0^xf\left(x,u\right)\:\right)\:$
but I wasn't sure if this was even right, or how to proceed from here:
 A: The key here isn't the Leibniz rule but the multivariable chain rule. Let
$$
 F(x) := \int_0^x \int_0^x f(s,t) \,ds\,dt,
$$
and observe that
$$
 F(x) = G(x,x), \quad G(u,v) := \int_0^v \int_0^u f(s,t) \,ds\,dt.
$$
In other words, $F(x) = G(u(x),v(x))$ for $u(x) = x$ and $v(x) = x$. Thus, by a first application of the multivariable chain rule,
$$
 F^\prime(x) = \frac{\partial G}{\partial u}\frac{d u}{d x} + \frac{\partial G}{\partial v}\frac{d v}{d x} = \frac{\partial G}{\partial u}(u(x),v(x)) + \frac{\partial G}{\partial v}(u(x),v(x)),
$$
and hence, by a second application of the multivariable chain rule,
$$
 F^{\prime\prime}(x) = \left(\frac{\partial}{\partial u}\left(\frac{\partial G}{\partial u}\right)\frac{du}{dx} +\frac{\partial}{\partial v}\left(\frac{\partial G}{\partial u}\right)\frac{dv}{dx} \right) + \left(\frac{\partial}{\partial u}\left(\frac{\partial G}{\partial v}\right)\frac{du}{dx} +\frac{\partial}{\partial v}\left(\frac{\partial G}{\partial v}\right)\frac{dv}{dx} \right)\\
= \frac{\partial^2 G}{\partial u^2}(u(x),v(x)) + \frac{\partial^2 G}{\partial v\partial u}(u(x),v(x)) + \frac{\partial^2 G}{\partial u \partial v}(u(x),v(x)) + \frac{\partial^2 G}{\partial v^2}(u(x),v(x))\\
= \frac{\partial^2 G}{\partial u^2}(x,x) + \frac{\partial^2 G}{\partial v\partial u}(x,x) + \frac{\partial^2 G}{\partial u \partial v}(x,x) + \frac{\partial^2 G}{\partial v^2}(x,x).
$$
At this point, all you need to do is compute the second-order partial derivatives of $G(u,v)$, but this is now a fairly straightforward exercise in applying the fundamental theorem of calculus, one variable at a time.
