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Theorem 1 (chain rule of Jacobian matrices).

Let $X = \mathbb R^m$, $Y = \mathbb R^n$, and $Z = \mathbb R^k$.

Let $f: D_f \subseteq X \to Y$ be differentiable at $\mathbf p \in D_f$.

Let $D_g \subseteq Y$ contains $f(\mathbf p)$, and let $g: D_g \subseteq Y \to Z$ be differentiable at $f(\mathbf p)$.

Then, $$ \mathbf J_{g \circ f}(\mathbf p) = \mathbf J_{g}(f(\mathbf p))\; \mathbf J_{f}(\mathbf p). $$

Wenchuan
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  • This is easily proved by using the linear approximation definition of the derivative. The proof works for maps between normed vector spaces, where partial differentiation is not enough to prove differentiability. – Mason Jun 09 '22 at 16:21
  • Thank you! What do you mean by "not enough"? (I noticed the issue of this proof in normed vector spaces, so I emphasis the spaces here are $\mathbb R^m$, $\mathbb R^n$, and $\mathbb R^k$, but not "a normed vector space over $\mathbb K$". But I am not sure if that is good enough) – Wenchuan Jun 09 '22 at 16:27
  • This is what I mean: https://math.stackexchange.com/questions/3745037/a-frech%C3%A9t-differentiable-function-to-has-a-continuous-derivative-in The method of using directional derivatives actually is not enough, even for $\mathbb{R}^n$ valued functions; there are functions where all directional derivatives exists, but the Frechet derivative doesn't. – Mason Jun 09 '22 at 17:43
  • So, should I prove $f \circ g$ is differentiable at $\mathbf p$ if $f$ and $g$ are differentiable at $\mathbf p$ first? – Wenchuan Jun 10 '22 at 01:02
  • What you wrote is the correct statement. But it isn't possible to prove it using partial derivatives since partial derivatives are not sufficient for differentiability. The simplest way to prove the statement is by expanding $$g(f(x + y)) = g(f(x) + Df(x)y + yR(y)) = \dots.$$ It makes the chain rule obvious. – Mason Jun 10 '22 at 02:03
  • Thank you! I do that again! – Wenchuan Jun 10 '22 at 02:15
  • MMM, yeah, I checked this again. In this proof I am trying to find what the Jacobian matrices is, but not the differentiability of the composite, as the differentiability is a pre-condition of the theorem. In your way, it is easy to prove that the chain rule of differentiability and is a good idea indeed. (Please be patient, I am not pro) – Wenchuan Jun 10 '22 at 02:31
  • Because, if the differentiability holds, then the partial derivatives exist, but, indeed, all partial derivatives exist dose not implies the differentiability. – Wenchuan Jun 10 '22 at 02:33
  • The connection between partial derivatives and differentiability is $\partial_i f(x) = Df(x)e_i$. The partial derivatives are the coordinates of $Df(x)$. – Mason Jun 10 '22 at 02:37
  • Hi Mason, I review the definition of $df$ I mentioned below. It is a differential but need not be a Jacobian matrix. You are right, this proof is not enough. Thank you! I delete the wrong answer I gave for now. It will be corrected! – Wenchuan Jun 12 '22 at 03:57
  • Yes, you're right, I can prove “the differential of the composite is the composite differential” by partial derivative. – Wenchuan Jun 12 '22 at 03:59

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