Let $T\in\mathcal D'(\mathbb R)$ be a distribution and let $f\in C^\infty_c(\mathbb R)$ be a test function.
How may I show that
$$(T*f)'=T'*f=T*f'\tag*{?}$$
This is what I did:
$$\begin{align*}(T*f)'(x)&=\lim_{h\to0}\frac{T*f(x+h)-T*f(x)}{h}\\&=\lim_{h\to0}\frac{\int T(y)f(x+h-y)-\int T(y)f(x-y)}{h}\\&=\lim_{h\to0}\frac{\int T(y)(f(x-y+h)-f(x-y))}{h}\\&=\int T(y)\lim_{h\to0}\frac{f(x-y+h)-f(x-y)}{h}\tag*{DCT}\\&=\int T(y)f'(x-y)\\&=T*f'(x)\end{align*}$$
Since convolution is commutative, one can similarly show that $(T*f)'=T'*f$.
Therefore, $?$ holds.
However, I feel that there is something off about my argument. For example, I did not use the fact that $f$ is compactly supported. Moreover, is $T$ not supposed to take arguments in $C_c^\infty(\mathbb R)$ and not in $\mathbb R$ since $T$ is a linear functional on $C_c^\infty(\mathbb R)$?