If $f,g\in \mathcal{S}$, then $(f\ast g)^{(n)} = f^{(n)}\ast g = f\ast g^{(n)}$ I want to prove that:

If $f,g\in \mathcal{S}$, then $(f\ast g)^{(n)} = f^{(n)}\ast g = f\ast g^{(n)}$

This is pretty much obvious if one knows that the differentiation can "enter the integral":
$$
     \frac{d}{dx} (f\ast g)(x) = \frac{d}{dx} \int f(x-y)g(y)\,dy = \int  \frac{\partial}{\partial x} f(x-y)g(y)\,dy
$$
(and then induction).
So this seems to be a job for the Leibniz integral theorem. However, I do not see how to apply it in this case.
 A: When convolving elements of $\mathscr S(\mathbb R)$ you are allowed to differentiate under the integral sign. Indeed
$$
\frac{1}{h}\int_{\mathbb R} \big(f(x+h-y)-f(x-y)\big)g(y)\,dy-\int_{\mathbb R} f'(x-y)g(y)\,dy\\=\frac{1}{h}\int_{\mathbb R} \big(f(x+h-y)-f(x-y)-hf'(x-y)\big)g(y)\,dy \\=\frac{1}{h}\int_{\mathbb R}\int_0^h\big(f'(x-y+t)-f'(x-y)\big)\,g(y)\,dy\,dt\\=\frac{1}{h}\int_{\mathbb R}\int_0^h \int_0^t f''(x-y+s) g(y)\,dy\,dt\,ds \\ =\frac{1}{h}\int_0^h \int_0^t (\,f''*g)(x+s)\,dt\,ds
$$
Clearly
$$
\left|\frac{1}{h}\int_0^h \int_0^t (\,f''*g)(x+s)\,dt\,ds\,\right|\le \|\,f''*g\|_\infty\cdot\frac{1}{h}\cdot h\cdot h=h\to 0.
$$
A: Integrate by parts:
\begin{align}
(f'*g)(x) & = \int_{-\infty}^\infty f'(y) g(x-y) \, dy = \int u \,dv = uv - \int v\,du \\
\text{where } & u = g(x-y) \text{ and } dv = f'(y)\,dy \\
\text{so that } & du = -g'(x-y) \, dy \text{ and } v = f(y) \\[12pt]
& =uv - \int v\,du = \left. g(x-y) f(y) \vphantom{\frac 1 1} \right|_{y\,;=\,-\infty}^\infty - \int_{-\infty}^\infty f(y)\Big(-g'(x-y)\,dy\Big) \\[10pt]
& = \int_{-\infty}^\infty f(y) g'(x-y)\, dy = (f*g')(x).
\end{align}
Postscript: I see that there is a bit more to the question than showing that $f'*g=f*g'.$ It was also desired to show that those are equal to $(f*g)'.$ If one knows about generalized functions such as "Dirac's" delta function $\delta,$ one can say that $\delta*f=f$ and then since $\delta'*f = \delta * f' = f',$ one can then say
$$
f'*g = (\delta'*f)*g = \delta'*(f*g) = (f*g)'.
$$
Without the theory of the delta function, there's further work to do.
