Consider the distribution
$$ f_a(x)=\frac{H(x+a)-H(x-a)}{2a}$$
Determine the $a$-derivative of this distribution
$$ \left < \frac{\partial f_a}{\partial a},\phi \right> = \lim_{h\to0} \frac{ \left <f_{a+h} ,\phi \right > - \left <f_{a} ,\phi \right > }{h}$$
and the limit
$$ \lim_{a\to 0} \frac{\partial f_a}{\partial a}$$
My partial answer can be found below.
We consider that we have $$ \langle f_{a+h},\phi\rangle-\langle f_a,\phi\rangle= \frac{1}{2(a+h)}\int_{-a-h}^{a+h} \phi(x)dx-\frac{1}{2a}\int_{-a}^{a} \phi(x)dx$$ Combining the fractions, $$ = \frac{1}{2a(a+h)}\left ( a\int_{-a-h}^{a+h} \phi(x)dx-(a+h)\int_{-a}^{a} \phi(x)dx \right) $$ Splitting the integrals,
$$ = \frac{1}{2a(a+h)}\left ( a\int_{-a-h}^{-a} \phi(x)dx +a\int_{-a}^{a} \phi(x)dx +a\int_{a}^{a+h} \phi(x)dx -a\int_{-a}^{a} \phi(x)dx-h\int_{-a}^{a} \phi(x)dx \right )$$
Cancelling, $$ = \frac{1}{2a(a+h)}\left ( a\int_{-a-h}^{-a} \phi(x)dx +a\int_{a}^{a+h} \phi(x)dx -h\int_{-a}^{a} \phi(x)dx \right )$$
So if we divide by $h$ and let $h\to 0$, we get $\frac{0}{0}$, so we may use L'Hopital's rule to get
$$ \left <\frac{\partial f_a}{\partial a},\phi\right >=\lim_{h\to 0} 2(a+2h)^{-1}a^{-1}\left ( a \phi(-a-h)+ a\phi(a+h)-\int_{-a}^{a} \phi(x)dx \right ) $$
Yielding
$$ \left <\frac{\partial f_a}{\partial a},\phi\right >= \frac{ a \phi(-a)+ a\phi(a)-\int_{-a}^{a} \phi(x)dx }{2a^2} $$
or
$$\frac{\partial f_a}{\partial a}= \frac{ a\delta(x+a)+a\delta(x-a) -2af_a }{2a^2} $$
That concludes part 1. Intuitively, I would have expected $-\delta(x-a)$, but this is what I have. Did I miss a negative sign?
So now what is the limit as $a\to 0$?
$$\lim_{a\to0 }\frac{\partial f_a}{\partial a}= \lim_{a\to0 }\frac{ a \phi(-a)+ a\phi(a)-\int_{-a}^{a} \phi(x)dx }{2a^2} $$
Using L'Hopital's rule again,
$$= \lim_{a\to0 }\frac{\phi(-a)+a\phi'(-a)(-1) + \phi(a)+a\phi'(a)-\phi(a)+\phi(-a)(-1) }{4a} $$
Cancelling,
$$= \lim_{a\to0 }\frac{a\phi'(-a)(-1) +a\phi'(a) }{4a} $$
or
$$= \lim_{a\to0 }\frac{\phi'(a) -\phi'(-a)}{4}=\frac{\phi'(0) -\phi'(0)}{4}=0 $$
Intuitively, I would have expected $\delta'(x)$, but this is the answer I have. Do you agree?