Currently, I am studying the Laplace transforms. I know that $X(p) = p$ cannot match with any $x(t)$ function, since $X(p) \notin L^1$. However, I decided to try computing the reverse Laplace transform to see what it gives.
I use the following properties:
$$\delta (t) \overset{\mathcal{L}}{\longleftrightarrow} 1 \tag{1}$$
$$\frac{d}{dt}x_0(t) \overset{\mathcal{L}}{\longleftrightarrow} pX_0(p) \tag 2$$
Using $X_0(p) = 1$, I deduce:
$$\color{red}{\frac{d}{dt}\delta(t)} \overset{\mathcal{L}}{\longleftrightarrow} p \tag 3$$
That's why I wonder: what is the derivative of the Dirac impulse ? It is, perhaps, a "forbidden" generalised function?
There is nothing special or "forbidden" about derivative of Dirac delta or any other generalized function. By definition, derivative $\varphi'$ of some generalized function $\varphi$ is a generalized function that satisfies $$ \left<\varphi', f\right> = \int_{\mathbb{R}} \varphi'(t) f(t)\mathrm{d}t = - \int_{\mathbb{R}} \varphi(t) f'(t)\mathrm{d}t = -\left<\varphi, f'\right> $$ for all test functions $f$.
Specifically, $\delta'$ is defined as $\left<\varphi', f\right> = -f'(0)$.