The one technique of differentiation in first-year calculus that is not introduced until after integrals are mentioned is this: $$ \frac d {dx} \int_a^x f(u) \, du = f(x). \tag 1 $$
I am failing to see how to show the following without using anything other than $(1){:}$ $$ \frac d {dx} \int_0^x \left( \int_0^{x-u} f(u)g(v) \, dv \right) \, du = \int_0^x f(x-v) g(v) \, dv. \tag 2 $$ One can write $$ \begin{bmatrix} s \\ t \end{bmatrix} = \begin{bmatrix} u+v \\ v \end{bmatrix}, \qquad \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} s-t \\ t \end{bmatrix} $$ and then $$ d(u,v) = \left|\frac{\partial(u,v)}{\partial(s,t)} \right| \, d(s,t) = 1 \, d(s,t) \tag 3 $$ and \begin{align} & \iint\limits_{u,v\,:\,0\,\le\,u,\,0\,\le\,v,\, u+v\,\le\,x} f(u)g(v)\, d(u,v) \\[10pt] = {} & \iint\limits_{s,t\,:\, 0\,\le\,t\,\le\,s\,\le\,x} f(s-t) g(t) \, d(s,t) \\[10pt] = {} & \int_0^x \left( \int_0^s f(s-t)g(t)\, dt \right) \, ds \end{align} and then apply $(1),$ getting $(2).$
However, I would prefer using only a one-variable chain rule rather than $(3),$ or better still, no chain rules. Is there a way to do that?
Here is a rough proof, some details would need to be written properly but it does work in the end.
Put $$F(x)= \int_0^x \left( \int_0^{x-u} f(u)g(v) \, dv \right) \, du $$
Note that $F(x)$ is the (signed) volume between the plane $z=0$ and the graph of the map $\Phi:\mathbb{R}^2\to \mathbb{R}: (u,v)\mapsto f(u)g(v)$.
Now
$$\frac{\partial F}{\partial x}=\lim_\limits{\varepsilon\to 0}\frac{F(x+\varepsilon)-F(x)}{\varepsilon}$$
When $\varepsilon$ is close to $0$, the difference $F(x+\varepsilon)-F(x)$ is the (signed) volume of a region in space lying over an isoceles trapezoid, of width $\frac{\varepsilon}{\sqrt{2}}$, ridiculously close to a rectangle of the same width, and length $x\sqrt{2}$.
The region you are looking at is delimited by that rectangle, vertical planes, and the graph of $\Phi$.
By applying a linear squeezing map $(s,t)\mapsto(s\sqrt{2},\frac{t}{\sqrt{2}})$ where $s$ and $t$ are coordinates adapted to the directions of the rectangle, you don't $(\star)$ change the volume of the region, but now the width of the rectangle is $\varepsilon$ and its length is $x$.
Hence the volume is equivalent as $\varepsilon\to 0$ to $$\varepsilon\int_0^x f(x-v) g(v) \, dv.$$
After dividing by $\varepsilon$ you get the expected limit.
$(\star)$ if you neglect the difference in height of the graph of $\Phi$ on either side of the $\varepsilon$ width.