Change of variables for step function before differentiating

Question

I have some confusion regarding the Heaviside step function and how and when variables can be changed. The issue comes from considering the function $H(e^x - e^{x_0})$. Taking the derivative with respect to $x$ gives $\frac{d}{dx} H(e^{x} - e^{x_0}) = e^x \, \delta(e^x - e^{x_0})$. However if you take the derivative of the function $H(x - x_0)$ you get $\frac{d}{dx} H(x - x_0) = \delta(x - x_0)$. The two step functions appear to be the same, but their derivatives aren't equivalent since using the composition rule for the Dirac delta function gives $ e^x \, \delta (e^x - e^{x_0}) = \frac{e^x}{e^{x_0}} \delta(x - x_0)$. Clearly something is going wrong here, but where?

Answer 1

If $g$ is smooth, we have (if chain rule is applicable for distributions) $$ \frac{d}{dx} H(g(x)) = g'(x) \delta(g(x)) = g'(x) \sum_{x' \text{ s.t. }g(x')=0} \frac{1}{|g'(x')|} \delta(x-x') . $$

For $g(x)=e^{x}-e^{x_0}$ we then have $g'(x) = e^{x}$ and $g(x')=0$ for $x'=x_0.$ Thus, $$ g'(x) \sum_{x' \text{ s.t. }g(x')=0} \frac{1}{|g'(x')|} \delta(x-x') = e^{x} \frac{1}{e^{x_0}} \delta(x-x_0) = e^{x_0} \frac{1}{e^{x_0}} \delta(x-x_0) = \delta(x-x_0) . $$

Thus, $$ \frac{d}{dx} H(e^{x}-e^{x_0}) = \frac{d}{dx} H(x-x_0) . $$