Question: Let $B\sim$ Bernoulli$(p)$ be a random variable drawn from a Bernoulli distribution with parameter $p$. What is the derivative $$\frac{d}{dp}B\:?$$
Background: I encountered this term while deriving the log-likelihood function of a specific distribution with the objective of finding the maximum likelihood estimate for $p$. There are many other terms in the original equation, but I am unsure about the derivative of this one.
What I tried so far: Since the cumulative distribution function is $$F_B(\omega)=\begin{cases}0\quad&\omega < 0\\1-p \quad& 0\le\omega< 1\\1 \quad &\omega \ge 1 \end{cases}$$ I would expect that the derivative is $$\frac{d}{dp}B = \begin{cases}-1 \quad& 0\le\omega< 1\\0 \quad &\text{else}\end{cases}$$ but I am not sure if this is correct because I don't know if I am allowed to simply plug in the CDF. Does this make sense?
If we think of $B$ as a function from some sample space $\Omega$ to $\{0,1\}$, then $B_p(\omega)$ is the value of $B$ with parameter $p$ at sample space point $\omega$.
Let's pick $\Omega = [0,1]$ and $B_p(\omega)=1_{>p}(\omega),;\ p \in [0,1]$ with probability measure being the 1-Lebesgue measure $\mu_1$ and the $\sigma-$field being $\mathcal{B}([0,1])$ so we get the probability space $S_0:=([0,1],\mathcal{B}([0,1]),\mu_1)$
With this, we see that $B_p(\omega)$ is a step function as a function of $p$, where $B_p(\omega)=1$ for $p<\omega$ and $0$ otherwise.
$$\frac{d}{dp}B_p(\omega) = \delta_p(\omega) $$
Where $\delta_p$ is the dirac delta function at $p$.