It seems to me that the derivative of the Heaviside function with a function as its argument should be via the chain rule:
$$\theta_{x}(f(x))=\delta(f(x))f_{x}(x) $$
But there is also the Delta distribution identity:
$$\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{f_{x}(a_{i})}\right|}$$ Where $a_{i}$ are the roots of the function f(x). This leads to the following equation:
$$\theta_{x}(f(x))=\sum_{i}\frac{\delta(x-a_{i})}{\left|{f_{x}(a_{i})}\right|}f_{x}(x)$$
which I do not understand. It does not seem to have a reasonable interpretation.
My question is: what is the derivative of a Heaviside function with a function as its argument?
Note, $f(x) \, \delta(x-a) = f(a) \, \delta(x-a)$, so $$ \theta_{x} (f(x)) = \sum_{i} \frac{\delta(x-a_{i})}{\left|{f_{x}(a_{i})}\right|}f_{x}(x) = \sum_{i} \frac{\delta(x-a_{i})}{\left|{f_{x}(a_{i})}\right|}f_{x}(a) = \sum_{i} \frac{f_{x}(a)}{\left|{f_{x}(a_{i})}\right|}\delta(x-a_{i}) \\ = \sum_{i} \operatorname{sign}(f_{x}(a_{i})) \, \delta(x-a_{i}) $$
Going to the definitions:
By definition, $$ \left\langle \frac{d}{dx} \theta(f(x)), \phi(x) \right\rangle = - \langle \theta(f(x)), \phi'(x) \rangle = - \int \theta(f(x)) \, \phi'(x) \, dx = - \int_{\{x \mid f(x)>0\}} \phi'(x) \, dx . $$
Now assume that $f>0$ on the nonoverlapping intervals $(a_i, b_i)$. Then we have $$ -\int_{\{x \mid f(x)>0\}} \phi'(x) \, dx = -\int_{\bigcup_i (a_i,b_i)} \phi'(x) \, dx = -\sum_i \int_{(a_i,b_i)} \phi'(x) \, dx = -\sum_i \left( \phi(b_i) - \phi(a_i) \right) \\ = -\sum_i \left( \langle \delta(x-b_i), \phi(x) \rangle - \langle \delta(x-a_i), \phi(x) \rangle \right) = \langle \sum_i \delta(x-a_i) - \sum_i \delta(x-b_i), \phi(x) \rangle . $$
Thus, $$ \frac{d}{dx} \theta(f(x)) = \sum_i \delta(x-a_i) - \sum_i \delta(x-b_i). $$
Note that if $f$ is differentiable, and thus continuous, then $a_i$ and $b_i$ are zeroes of $f$, and if $f_x \neq 0$ at $a_i$ and $b_i$ then $f_x(a_i) > 0$ and $f_x(b_i) < 0$, so $\operatorname{sign}(f_x(a_i)) = +1$ and $\operatorname{sign}(f_x(b_i)) = -1$.
Thus, $$ \frac{d}{dx} \theta(f(x)) = \sum_{c_i = a_i, b_i} \operatorname{sign}(f_x(c_i)) \, \delta(x-c_i) $$ in accordence with the equation at the top.