Derivative of a one-parameter dependent distribution

Question

Below, the notation $\max(t,0)\equiv \max(t)$ is used, and the function $\max$ is seen as a distribution. From the theory of distributions, we have that $\max'(t)=H(t)$ (in the distributional sense) where $H$ is the Heaviside distribution. Now, I am considering the one-parameter distribution $\max(at)$ where $a$ is the real parameter. I am interested in the derivative of $\max(at)$, seen as a distribution in $t$, with respect to the parameter $a$. Here is what I think is more or less correct but things are a bit vague: using the integral notation for distributions (makes sense because the distribution in question is a classical integrable function) $$\forall \phi,\quad \frac{d}{da}\Bigl(\int \phi(t)\max(at)dt\Bigr)=\int\phi(t)\frac{d}{da}(\max(at))dt=\int\phi(t)t \max'(at)dt=\int\phi(t)t H(at)dt $$ where the classical differentiation rule for composite functions, that is $\tfrac{d}{da}f(ax)=xf'(ax)$ has been used in a distributional sense on $\max(ax)$ (not sure it is acceptable). In other words, in the sense of distributions, $$ \frac{d}{da}(\max(at))=t H(at) $$ Is it correct in general to proceed in this way?

Answer 1

If $a>0$ we have $\max(at)=a\max(t)$ so $$\frac{d}{da}\max(at) = \max(t).$$

If $a<0$ we have $\max(at)=\max(-|a|t)=-\min(|a|t)=-|a|\min(t)=a\min(t),$ where $\min(t):=\min(t,0),$ so $$\frac{d}{da}\max(at) = \min(t).$$