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If $\mathbf f : \mathbb R^n \to \mathbb R^m$ is differentiable at $x_0 \in \mathbb R^n$, define the differential operator $$d\mathbf f_{x_0} : \mathbb R^n \to \mathbb R^m $$ as the linear operator that takes a small increment $h \in \mathbb R^n$ and outputs $D\mathbf f(x_0) \cdot h$, where $D\mathbf f(x_0) \in \operatorname {Mat}(m\times n,\mathbb R)$ is the Jacobian matrix of $\mathbf f$ at the point $x_0$.

Let $\mathbf f$ be defined as before, and let $\mathbf g : \mathbb R^p \to \mathbb R^q$ be differentiable at $y_0 \in \mathbb R^p$. Call $a_0 \doteq \mathbf f(x_0)$ and $b_0 \doteq \mathbf g(y_0)$. Lastly, let $\mathbf h : \mathbb R^m \times \mathbb R^q \to \mathbb R^s$ be differentiable at $(a_0,b_0)$. How would I write down the differential $d\mathbf s_{(x_0,y_0)}$ of function $\mathbf s : \mathbb R^n \times \mathbb R^p \to \mathbb R^s$ defined as follows, $$\mathbf s : (x,y) \mapsto \mathbf h(\mathbf f(x),\mathbf g(y)) $$ as a composition of the differentials $d\mathbf h_{(a_0,b_0)}$, $d\mathbf f_{x_0}$, and $d\mathbf g_{y_0}$?

I know that the Jacobian matrix associated to $d\mathbf s_{(x_0,y_0)}$ should be in $\operatorname{Mat}(s \times pq, \mathbb R)$, but I'm not exactly sure what matrices need be multiplied together to construct it. I guess my brain is having a hard time understanding how to "merge" two functions such as $\mathbf f$ and $\mathbf g$ in an ordered couple, so I can't compile my composition diagram correctly.

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    if $f: \mathbb{R}^p \rightarrow R^m$, $g:R^m \rightarrow R^n$ are two differentiable functions respectively at point $a\in \mathbb{R}^p$ and at point $f(a)$ then for $h\in \mathbb{R}^p$, $d (g\circ f)_a(h) = d g_{f(a)} (df_a(h))$. – 2017-02-10

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$$d\mathbf{s}_{(x_0,y_0)}=d\mathbf{h}_{(a_0,b_0)}\cdot(d\mathbf f_{x_0}+d\mathbf g_{y_0})$$

Where $d\mathbf f$ and $d\mathbf g$ are viewed as elements of $\mathrm{M}_{(m+q)\times(n+p)}$ in the "obvious" way: $$d\mathbf f \equiv \begin{pmatrix}d\mathbf f_{m\times n}&0_{q\times n }\\ 0_{m\times p}&0_{q\times p}\end{pmatrix},\qquad d\mathbf g \equiv \begin{pmatrix}0_{m\times n}&0_{q\times n }\\ 0_{m\times p}&d\mathbf{g}_{q\times p}\end{pmatrix}$$ So $$d\mathbf f+d\mathbf g\equiv\begin{pmatrix}d\mathbf f_{m\times n}&0_{q\times n }\\ 0_{m\times p}&d\mathbf g_{q\times p}\end{pmatrix}$$

This follows from the chain rule together with $(\mathbf f,\mathbf g):\Bbb R^n\times\Bbb R^p\to\Bbb R^m\times \Bbb R^q$ having derivative $d\mathbf f+d\mathbf g$ viewed as above. That can be seen by calculating for example the components of the derivative explicitly.