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In my Introduction to Differential Multivariable Calculus class we have proven the Chain Rule theorem, which says:

Given an open set $D$ of $\mathbb{R}^n$, $a \in D$, and $f:D \to \mathbb{R}^m$ a diferentiable function in $a$. Suppose that $f(D) \subset E$ is another open, and that $g:E \to \mathbb{R}^p$ is differentiable at $g(a)$. Then the composition is differentiable at a, and it follows that $D(g \circ f)(a) = D(g(f(a)) \circ D(f(a))$, where $D(F)$ is the Jacobian matrix of the function $F$.

Now, my question is about the notation used. In the theorem of the Existence Of the Implicit Function, there's a use of the Chain Rule that I can't understant: given $F(g(x),x)=0$, where $F:\mathbb{R}^n \times \mathbb{R}^p \to \mathbb{R}^n$ (the function is not important here), we can obtain because of the Chain Rule the following relation: $$D_y(F(g(x),x))D(g(x)) + D_zF(g(x),x))=0,$$ where $D_yF= \frac{\delta(F_1,\cdots,F_n)}{\delta(y_1,\cdots,y_n)}$ is the jacobian matrix, and same for $D_zF$ ($y,z$ are vectors of $\mathbb{R}^n,\mathbb{R}^p$).

I tried to apply the theorem and the given notation, and I attempted to multiply the Jacobians, but I think I'm getting lost in the notation used.

Can someone explain in detail how the equation is derived from the Chain rule?

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    $\begingroup$ View the map $x\mapsto {g(x)\choose x}$ as going from $\mathbb R^p$ to $\mathbb R^n\times\mathbb R^p\,.$ Its Jacobian is a block matrix that you multiply with the Jacobian of $F\,.$ $\endgroup$ Commented Sep 12, 2024 at 15:05
  • $\begingroup$ That's what I thought, because then it would made sense that there's the Identity being multiplied. However, if $x \to x$ is the identity, then we could decompose any jacobian matrix with $D_y + D_z$, but that doesn't make any sense to me. They aren't event the same dimension, right? $\endgroup$ Commented Sep 13, 2024 at 10:17
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    $\begingroup$ Do not mechanically manipulate symbols. Come up with a low dimensional concrete example to see the matrices in action. $\endgroup$ Commented Sep 13, 2024 at 10:19
  • $\begingroup$ In implicit function case that prompted you to set up this post we don't calculate the Jacobian of $F(id(x),id(x))\,.$ Go back to my first comment and restart. $\endgroup$ Commented Sep 13, 2024 at 10:34
  • $\begingroup$ Then, let $h(x): \mathbb{R}^p \to \mathbb{R}^{n+p}$, by $(g(x),x)$. Then $D(F \circ h)= D(F \circ h)D(h)$. But $D(h)= \begin{pmatrix} D(g(x)) \\ D(x) \end{pmatrix}$, and $D(F \circ h)= (\frac{\delta F_1, \cdots, F_n}{\delta x_1,\cdots,x_n},\frac{\delta F_1, \cdots, F_n}{\delta x_{n+1},\cdots,x_p})$. Then we multiply the block matrixes and we get the result. Thanks $\endgroup$ Commented Sep 13, 2024 at 10:49

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You should mention that the arguments of $F : \mathbb R^n \times \mathbb R^p \to \mathbb R^n$is written in the form $(y,x)$ with $y \in \mathbb R^n, x \in \mathbb R^p$. Then $DF$ can be split into the two submatrices $$D_yF = \begin{pmatrix} \frac{\partial F_1}{\partial y_1} & \ldots & \frac{\partial F_1}{\partial y_p} \\ \vdots & & \vdots \\ \frac{\partial F_n}{\partial y_1} & \ldots & \frac{\partial F_n}{\partial y_p} \\ \end{pmatrix}$$ $$D_xF = \begin{pmatrix} \frac{\partial F_1}{\partial x_1} & \ldots & \frac{\partial F_1}{\partial x_n} \\ \vdots & & \vdots \\ \frac{\partial F_n}{\partial x_1} & \ldots & \frac{\partial F_n}{\partial x_n} \\ \end{pmatrix}$$ $$DF = \begin{pmatrix}D_yF & D_xF \end{pmatrix}.$$ Now you have a function $g : \mathbb R^n \to \mathbb R^p$ and define $$G : \mathbb R^n \to \mathbb R^p \times \mathbb R^n, G(x)= (g(x),x) = (g(x),id(x)).$$ Then $DG$ can be split into the two submatrices $Dg$ and $Did = I_n =$ $(n \times n)$-unit matrix:

$$DG = \begin{pmatrix} Dg \\ I_n \end{pmatrix}$$ Hence $$D(F \circ G)(x) = DF(G(x)) \circ DG(x) = D_yF(G(x)) \circ Dg(x) + D_xF(G(x)) \circ I_n \\= D_yF(G(x)) \circ Dg(x) + D_xF(G(x)) \\= D_yF(g(x),x)) \circ Dg(x) + D_xF(g(x),x)) $$ Since $F \circ G = 0$, we get the equation $$D_yF(g(x),x)) \circ Dg(x) + D_xF(g(x),x)) = 0 .$$

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