Delta method with mix of continuous and discrete variables

This is my first question on Cross Validated so please bear with me if my question is lagging in any dimension. My question regards how to evaluate a Jacobian matrix when one variable is binary. I have googled quite extensively but found no answer.

Setup:

I have three variables $p$, $r$ and $s$. $p$ and $r$ come from a copula and thus have uniform marginal distributions and covariance $E[(p-\bar p)(r-\bar p )]=\sigma_{pr}$. The binary variable $s\in\{0,1\}$ is independent of $p$ and $r$ and $E[s]=w$.

I define the vector $\mathbf{x}$ with mean $\mathbf{\bar x}$ and variance$\mathbf{\Sigma}_\mathbf{x}$ : $$\mathbf{x}=\begin{pmatrix} p\\ r\\ s \end{pmatrix}, \mathbf{\bar x}=\begin{pmatrix} \bar p\\ \bar r\\ w \end{pmatrix},\mathbf{\Sigma}_\mathbf{x}=\begin{pmatrix} \sigma_p^2&\sigma_{pr}&0\\ \sigma_{pr}&\sigma_p^2&0\\ 0 &0&w(1-w) \end{pmatrix}$$

The variables $r$ and $s$ are used to construct a new variable with the function $\Lambda(r,s)$ defined as: $$\Lambda(r,s)=wG\left(s{G}^{-1}(r)+(1-s){F}^{-1}(r)\right)+(1-w)F\left(s{G}^{-1}(r)+(1-s){F}^{-1}(r)\right)$$ $F$ and $G$ are continuous distributions with support on the positive real line.

I want to know the covariance between $p$ and the new variable created by $\Lambda(r,s)$. My approach has been to use the Delta method.

I define a new vector: $$\mathbf{y}=q(\mathbf{x})=\begin{pmatrix} p\\ \Lambda(r,s) \end{pmatrix}$$ In the case where all variables where continuous I would use the delta method to find an expression for the covariance matrix of $\mathbf{y}$: $$\mathbf{\Sigma}_\mathbf{y}=\mathbf{D}\mathbf{\Sigma}_\mathbf{x}\mathbf{D}^T$$ where $\mathbf{D}$ is the Jacobian evaluated at $\mathbf{\bar x}$, $\mathbf{D}=\mathbf{J}\Big|_{\mathbf{x}=\mathbf{\bar x}}$.

Question

The last part is my central issue: how (if possible) can I calculate $\mathbf{D}$ when $s$ is a binary variable?

My own Suggestion

I have been toying around with central differences. But I don't know if this approach works for the delta method.

The Jacobian of $q(\mathbf{x})$ in this case could be defined as: $$\mathbf{J}=\begin{pmatrix} 1&0 &0\\ 0 &\frac{\partial\Lambda(r,s)}{\partial r} &\frac{\Lambda(r,1)-\Lambda(r,0)}{2} \end{pmatrix}$$ But how do I get from $\mathbf{J}$ to $\mathbf{D}$? It is not obvious to me how to evaluate the $\frac{\partial\Lambda(r,s)}{\partial r}$ and $\frac{\Lambda(r,1)-\Lambda(r,0)}{2}$, which is necessary for knowing $\mathbf{D}$.