Is this mathematically correct for a basic proof that the derivative of the Heaviside function is equal to the delta function? I don't know much about distributions so I kept everything integrated.
$$H(x) = \begin{cases}0 & x < 0\\\frac{1}{2} & x = 0\\1 & x>0\end{cases}$$ $$\delta (x) = \begin{cases}\infty & x = 0\\0 & x\neq 0\end{cases}$$
$$\forall a > 0$$
$$\int_{-\infty}^{\infty}\delta (x) \;dx = \int_{-\infty}^{a}\delta (x) \;dx = 1$$
$$\frac{d}{da}\int_{-\infty}^{a}H(x)\;dx = \frac{d}{da}\int_{0}^aH(x)\;dx = \frac{d}{da}\int_{0}^adx = 1$$
$$\therefore\frac{d}{da}\int_{-\infty}^{a}H(x)\;dx = \int_{-\infty}^{a}\delta (x) \;dx $$
A distribution is a linear functional on a space of "nice" functions, e.g. $C_c^\infty$, infinitely differentiable functions with compact support. The application of a distribution $u$ on a nice function $\phi$ is often written $\langle u, \phi \rangle$.
The derivative of a distribution is defined by $\langle u', \phi \rangle := - \langle u, \phi' \rangle$. This formula is motivated by integration by parts, $\int f'(x) \, \phi(x) \, dx = - \int f(x) \, \phi'(x) \, dx$ when $\phi(x) = 0$ for big $|x|$.
These definitions give, for any smooth $\phi$ which vanishes at infinity, $$ \langle H', \phi \rangle = - \langle H, \phi' \rangle = - \int H(x) \, \phi'(x) \, dx = - \int_0^\infty \phi'(x) \, dx \\ = - \left( \phi(\infty) - \phi(0) \right) = \phi(0) = \langle \delta, \phi \rangle .$$
Since this is valid for all nice $\phi$ we have $H' = \delta$.