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In the single variable calculus, when doing a u substitution, we have $\int_a^b{f'(x)}dx = \int_a^b{(h \circ g)'(x)dx} = \int_a^b{(h' \circ g)(x)g'(x)dx} = (h \circ g)(x) \Big|_a^b = h(g(b)) - h(g(a)) = \int_{g(a)}^{g(b)}{h'(u)du}$

Here, there's a nice connection between the chain rule and the Jacobian, where the $g'(x)dx$ term comes out of the wash without any appeals to geometric reasoning, like is often the case when introducing the Jacobian term for the multivariable change of variables. Is it possible to make an analogous derivation for the multivariable case?

What confuses me is that the multivariable chain rule has the form of a sum, but the Jacobian term has the form of a difference, so at first glance it seems not possible.

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    $\begingroup$ “the multivariable chain rule has the form of a sum” Dear child, the multivariable chain rule is best written in matrix notation: if $f = g \circ h$ then $f’(x) = g’(h(x)) h’(x)$. Here $h: \mathbb R^n \to \mathbb R^m$ is differentiable at $x$, $g:\mathbb R^m \to \mathbb R^k$ is differentiable at $h(x)$ and $h’(x)$ is an $m \times n$ matrix and $g’(h(x))$ is a $k \times m$ matrix. “The Jacobian has the form of a difference” The Jacobian determinant is the determinant of a particular derivative matrix! $\endgroup$
    – littleO
    Aug 3 at 20:52
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    $\begingroup$ Relevant thread about Jacobian matrix: math.stackexchange.com/a/1127350/40119 Relevant post about change of variables theorem and Jacobian determinant: math.stackexchange.com/a/464972/40119 $\endgroup$
    – littleO
    Aug 3 at 21:07
  • $\begingroup$ There needs to be an integration by parts somewhere in the derivation, that's where something in the form of a difference will come from. $\endgroup$ Aug 4 at 19:54
  • $\begingroup$ Here’s a fundamental fact: Suppose that $A$ is a real $n \times n$ matrix, $b \in \mathbb R^n$ and $T:\mathbb R^n \to \mathbb R^n$ is defined by $T(x) = Ax + b$ for all $x \in \mathbb R^n$. If a set $S \subset \mathbb R^n$ has volume (or Lebesgue measure) $V$, then the volume (or Lebesgue measure) of $T(S)$ is $| \det A | V$. That is where the Jacobian determinant comes from in the change of variables formula for integration. More details are given in the posts in linked above. $\endgroup$
    – littleO
    Aug 4 at 20:12
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    $\begingroup$ I took a look at the article you mentioned, and the proof there is nothing like the 1D case (the article is very technical, as expected). The typical proof in the 1D case is rather special in that it exploits the FTC and the chain rule. By using the FTC, one by-passes a lot of the geometry regarding how a mapping $g$ distorts lengths/volumes. Also, 1D is trivial because $g$ maps intervals to intervals. The above links by littleO are, I think, the best ways to think about the theorem: always start with the linear case, and understand how determinants relate to volumes, and then generalize. $\endgroup$
    – peek-a-boo
    Aug 5 at 21:21

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