Does the function $u_{\alpha}(x)$ has a canonical name? What is the derivative of it at $\dfrac{\alpha}{2}$?

Question

I am learning this post to understand this claim

the derivative of a step function is the Dirac delta.

Consider the unit step function u(x) defined by

\begin{equation} \hspace{50pt} u(x) = \left\{ \begin{array}{l l} 1 & \quad x \geq 0 \\ 0 & \quad \text{otherwise} \end{array} \right. \hspace{50pt} (4.8) \end{equation}

This function has a jump at x=0. Let us remove the jump and define, for any α>0, the function $u_α$ as \begin{equation} \nonumber u_{\alpha}(x) = \left\{ \begin{array}{l l} 1 & \quad x > \frac{\alpha}{2} \\ \frac{1}{\alpha} (x+\frac{\alpha}{2}) & \quad -\frac{\alpha}{2} \leq x \leq \frac{\alpha}{2} \\ 0 & \quad x < -\frac{\alpha}{2} \end{array} \right. \end{equation}

$u_{\alpha}(x)$ is a continuous function. Now let us define the function $\delta_{\alpha}(x)$ as the derivative of $u_{\alpha}(x)$ wherever it exists.

\begin{equation} \nonumber \delta_{\alpha}(x)=\frac{ d u_{\alpha}(x)}{dx} = \left\{ \begin{array}{l l} \frac{1}{\alpha} & \quad |x| < \frac{\alpha}{2} \\ 0 & \quad |x| > \frac{\alpha}{2} \end{array} \right. \end{equation}

question

1. Does the function $u_{\alpha}(x)$ has a canonical name?

2. Is the function above not differentiable at $\dfrac{\alpha}{2}$

or

It doesn't matter what the derivative of $u_{\alpha}(x)$ at $\dfrac{\alpha}{2}$ is

or

the derivative of $u_{\alpha}(x)$ at $\dfrac{\alpha}{2}$ is also $\dfrac{1}{a}$ ?