Proof that Differentiating the Heaviside Step Function Yields the Dirac Delta Function
Theorem
The derivative of the Heaviside step function is the Dirac delta function.
$$ \dfrac{dH}{dx}=\delta (x) $$
Here, $H=H(x)$ refers to the Heaviside step function or unit step function
$$ H(x)=\begin{cases} 1 & x>0 \\ 0 & x \le 0 \end{cases} $$
A function that satisfies the following two conditions is called the Dirac delta function.
$$ \begin{equation} \delta (x) = \begin{cases} 0, & x\neq 0 \\ \infty , & x=0 \end{cases} \label{condition1} \end{equation} $$
$$ \begin{equation} \int_{-\infty}^{\infty}{\delta (x) dx}=1 \label{condition2} \end{equation} $$
Proof
We prove $\dfrac{dH}{dx}$ is the Dirac delta function by checking if it satisfies the two required conditions.
Condition $\eqref{condition1}$
Since $H(x)$ is a constant function in $x \neq 0$, it is $\dfrac{dH}{dx}=0$, and in $x=0$, the tangent line is a vertical line parallel to the $y$ axis, the derivative diverges to $\dfrac{dH}{dx}=\infty$. Therefore, $$ \dfrac{dH}{dx} = \begin{cases} \infty & x=0 \\ 0 & x \neq 0 \end{cases} $$
Condition $\eqref{condition2}$
$$ \begin{align*} \int _{-\infty} ^{\infty} \dfrac{dH}{dx} dx &= \int_{-\infty} ^{\infty} dH \\ &= \left[ H \right]_{-\infty}^{\infty} \\ &= 1-0 =1 \end{align*} $$
Since $\dfrac{dH}{dx}$ satisfies both conditions required for the Dirac delta function, we obtain the following result.
$$ \dfrac{dH}{dx}=\delta (x) $$
■