Derivative of a step function

Question

The derivative of the unit step function is: $\frac{d \theta (x)}{dx} = \delta (x) .$

However, we also have that $\theta(cx) = \theta(x)$ where $c$ is some constant, and so $$\frac{d \theta (x)}{dx} = \frac{d \theta (cx)}{dx} = c\delta (x)$$

In general I am wondering how we can generalize the derivative of a step function.

Answer 1

Note that $\theta(cx)=\theta(x)$ for $c>0$. On differentiating we get$$\theta'(x)=\frac{d\theta(cx)}{d(cx)}\frac{d(cx)}{dx}=c\delta(cx)$$$c\delta(cx)$ is equal to $\delta(x)$ for $c>0$. (Why? Compare their values for $x=0$ and $x\ne0$).

Answer 2

A step function $f$ is a finite linear combination of indicator functions of intervals. Say we have intervals $(I_k)_{k\leqslant n}$ and scalars $(\lambda_k)_{k\leqslant n}$ such that $$f = \sum_{k\leqslant n} \lambda_k \cdot \mathbb{I}_{I_k} $$

On just use linearity of derivation. $f$ is differentiable at $x$ iff $\mathbb{I}_{I_k}$ is differentiable at $x$ for all $k$, and one then has $$f'(x) = \sum_{k\leqslant n} \lambda_k \cdot \mathbb{I}_{I_k}'(x)$$

But $\mathbb{I}_{I_k}$ is differentiable for all $x$ except for the (finite) extremities of $I_k$, and $\mathbb{I}_{I_k}'=0$. Btw in OP $\delta = 0$ except for $x=0$ and $x=1$ where it isn't defined.

Thus, $f'$ is identicaly $0$, except for the finitely manies extremities of the intervals $\mathbb{I}_{I_k}$ where it isn't defined.