As the title suggest, I am looking to understand which is the result of such operation. In fact, I am willing to find the root (numerically) of:
$$ \sqrt{(f(x)H(f(x)))^2+(g(x)H(g(x)))^2}-K=0 $$
The derivative of such thing is:
$$ \frac{1}{\sqrt{H(f(x))^2+H(g(x))^2}}\left[f(x) H(f(x)) \left(\frac{\partial f(x)}{\partial x}H(f(x))+f(x) \delta(x) \frac{\partial f(x)}{\partial x} \right)+ g(x) H(g(x)) \left(\frac{\partial g(x)}{\partial x}H(g(x))+g(x) \delta(x) \frac{\partial g(x)}{\partial x} \right)\right] $$
Now, the terms like $f^2(x) H(f(x)) \delta(x)$, for $x=0$ are not defined (or at least is what my engineering head tells me, being infinity times zero undetermined).
On the other side, if I approximate
$$ H(x) \approx \bar{H}(x,\epsilon)= \left\{ \begin{array}{l} 0 \quad x \le -\epsilon \\ \frac{x + \epsilon}{2 \epsilon} \quad -\epsilon<x<\epsilon \\ 1 \quad x \ge \epsilon \\ \end{array} \right. \quad \text{ and } \quad \delta(x) \approx \bar{\delta}(x,\epsilon)= \left\{ \begin{array}{l} 0 \quad x \le -\epsilon \\ \frac{1}{2 \epsilon} \quad -\epsilon<x<\epsilon \\ 0 \quad x \ge \epsilon \\ \end{array} \right. $$
Then I have
$$ \lim_{\epsilon \rightarrow 0} \bar{H}(0,\epsilon)\bar{\delta}(0,\epsilon)= \infty $$
However, this does not convince me at all because, as I saw here Relation between Heaviside step function to Dirac Delta function , the Dirac delta makes sense only in a integral sense.
How can I solve this ambiguity?