If $f(x)=g(x)$, then $f'(x)=g'(x)$, is it right?
Assume there is a function $f(x)=m(x)\delta(x)$，where $m(x)$ is any function and $\delta(x)$ is Dirac-δ function. We know that $m(x)\delta(x)=m(0)\delta(x)$, so we have $f(x)=m(0)\delta(x)$.
But why $[m(0)\delta(x)]'\ne[m(x)\delta(x)]'$? $[m(0)\delta(x)]'=m(0)\delta'(x)$, and $[m(x)\delta(x)]'=m'(0)\delta(x)+m(x)\delta'(x)$, and I think it's obviously not equal.
They are actually the same: $[m\delta]'=m(0)\delta'$ and $[m\delta]'=m'\delta+m\delta'$.
The first equality follows from the definition of the distributional derivative:
$$[m\delta]'(\phi)=-[m\delta](\phi')=-m(0)\phi'(0)=m(0)\delta'(\phi).$$
The second equality follows from the distributional product rule.
How can we reconcile these, when it seems like the first version does not involve $m'(0)$ while the second one does? We notice that the $m\delta'$ term in the second version has its own product rule:
$$[m\delta'](\phi)=\delta'[m\phi]=-m'(0)\phi(0)-m(0)\phi'(0).$$
The first term here serves to cancel out the first term arising in the distributional product rule (the one depending on $m'(0)$). In the end only the second term survives.