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I am currently learning multivariable calculus and the following was done in my lecture to prove one of the properties of Jacobian

The property:

($JJ'=1$), where $J$ is $\frac{\partial (x,y)}{\partial(u,v)}$ and $J'$ is $\frac{\partial(u,v)}{\partial (x,y)}$

The equation in the proof:

$\partial u = \frac{\partial u}{\partial x} \partial x +\frac{\partial u}{\partial y} {\partial y}$

Then the professoor proceeded to divide by $\partial u$ on both sides so you end up with $\frac{\partial u}{\partial u}$ on the left hand side which is essential 1. He then did this for $v$ also.

My question: $\partial u\ , \partial x\ , \partial y$ on their own makes no sense. And since partial derivative is defined for suppose u with respect to some other variable, dividing or multiplying by $\partial (some \ variable)$ should be wrong. What is the correct theory ?

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    $\begingroup$ Sounds like garbage to me. And no one writes this meaningless notation. Get a new professor. And be aware that $\frac{\partial x}{\partial u}\frac{\partial u}{\partial x} \ne 1$. $\endgroup$ – Ted Shifrin Jan 8 at 17:29
  • $\begingroup$ This is just linear algebra -- that is, show that the product of a matrix and its inverse is the identity matrix, which is trivially true by definition. $\endgroup$ – Allawonder Jan 8 at 21:33
  • $\begingroup$ @TedShifrin i thought so too. Sadly my university doesn't let me choose the professor as this is a compulsory first year course. $\endgroup$ – Shaurya Goyal Jan 10 at 17:15
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You are correct: the notation $\partial u$, etc, is not really clear. Dividing the equation in the question by $\partial u$ would give, formally, $$ \frac{\partial u}{\partial u}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial u} $$ which doesn't make much sense.


Here is how I would handle this:

The Chain Rule says $$ \left.\begin{align} \frac{\mathrm{d}u}{\mathrm{d}t}&=\frac{\partial u}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial u}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}\\[3pt] \frac{\mathrm{d}v}{\mathrm{d}t}&=\frac{\partial v}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial v}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} \end{align}\right\}\hspace{.5cm} \begin{bmatrix}\dfrac{\mathrm{d}u}{\mathrm{d}t}\\\dfrac{\mathrm{d}v}{\mathrm{d}t}\end{bmatrix}=\begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} \begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}\tag1 $$ and $$ \left.\begin{align} \frac{\mathrm{d}x}{\mathrm{d}t}&=\frac{\partial x}{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}+\frac{\partial x}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}t}\\[3pt] \frac{\mathrm{d}y}{\mathrm{d}t}&=\frac{\partial y}{\partial u}\frac{\mathrm{d}u}{\mathrm{d}t}+\frac{\partial y}{\partial v}\frac{\mathrm{d}v}{\mathrm{d}t} \end{align}\right\}\hspace{.5cm} \begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}=\begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} \begin{bmatrix}\dfrac{\mathrm{d}u}{\mathrm{d}t}\\\dfrac{\mathrm{d}v}{\mathrm{d}t}\end{bmatrix}\tag2 $$ Plugging $(1)$ into $(2)$ gives $$ \begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix} =\begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} \begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} \begin{bmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\\dfrac{\mathrm{d}y}{\mathrm{d}t}\end{bmatrix}\tag3 $$ That is, $$ \begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} \begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} =I\tag4 $$

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  • $\begingroup$ Same as Charles answer: I don't think you answer the main intention of the question. $\endgroup$ – ConvexHull Jan 8 at 17:15
  • $\begingroup$ @ConvexHull: You might say that Charles' answer was the same as mine (as mine was 3 hours earlier). In any case, the question was "What is the correct theory ?" I have added a bit of an introduction mentioning that the OP's concerns are well-founded, but my answer still stands. $\endgroup$ – robjohn Jan 8 at 18:12
  • $\begingroup$ The point is that i think both answers are correct, however don't fit the intention of the question. The questionier is confused by the use of the differential as mathematical object. $\endgroup$ – ConvexHull Jan 8 at 18:16
  • $\begingroup$ Dividing by a partial "differential" $\partial u$ makes no sense at all. $\endgroup$ – ConvexHull Jan 8 at 18:34
  • $\begingroup$ @ConvexHull: I agree. I have added "formally" as that was my intent. That is what I assumed the instructor had done. $\endgroup$ – robjohn Jan 8 at 18:37
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There's nothing too deep going on here. The point is that if $f : \mathbb{R}^n \to \mathbb{R}^n$ and $g : \mathbb{R}^n \to \mathbb{R}^n$ satisfy $f \circ g = I$, then $Df|_{g(x)}Dg|_{x} = DI|_{x} = I$. All we used was the chain rule and the fact that the Jacobian of the identity transformation is the identity matrix.

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  • $\begingroup$ I don't think you answer the main intention of the question. $\endgroup$ – ConvexHull Jan 8 at 17:14
  • $\begingroup$ I was trying to give the "correct theory," i.e. what I take to be the clearest way to think about the problem. But I take your point that I didn't address his concern about dividing by $\partial u$, which, as far as I know, is just not a legitimate operation. $\endgroup$ – Charles Hudgins Jan 8 at 21:55
  • $\begingroup$ I was going add in something about why $\partial u$ is nonsense, but it looks like your post covers it. $\endgroup$ – Charles Hudgins Jan 8 at 22:01
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First of all, it is important to distinguish between expressions such as $dx$ and $\partial x$.

The differential was first introduced via an intuitive or heuristic definition by Leibniz, who thought of the differential $dy$ as an infinitely small change in the value $y$ of the function, corresponding to an infinitely small change $dx$ in the function's argument $x$.

Later Cauchy's approach was a significant logical improvement over the infinitesimal approach of Leibniz because, instead of invoking the metaphysical notion of infinitesimals. That is, one was free to define the differential $dy$ by an expression

$$dy := f'(x) dx $$

in which $dy$$dx$ and $f'(x)$ are simply new variables taking finite real values, not fixed infinitesimals as they had been for Leibniz.

Summarizing this: Total differentials such as $dx$, $dy$ and partial derivatives $f'(x) \equiv \frac{\partial y}{\partial x}$ must be thought as mathematical objects and can be manipulated in exactly the same manner as any other real quantities in a meaningful way. Note that this does not hold for $\partial x$ or $\partial y$ alone, which have no meaning in Cauchy's approach.

This might help you:

Differential of a function

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    $\begingroup$ Provide context for links: Links to external resources are encouraged, but please add context around the link so your fellow users will have some idea what it is and why it’s there. Always quote the most relevant part of an important link, in case the external resource is unreachable or goes permanently offline. $\endgroup$ – robjohn Jan 8 at 11:11
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    $\begingroup$ The use of $\partial$ here is inappropriate. $\endgroup$ – Ted Shifrin Jan 8 at 17:30
  • $\begingroup$ @TedShifrin Sure, the onedimensional case is inappropriate in a logical sense. However, it is useful to describe the different interpretations and therefore has its justification. $\endgroup$ – ConvexHull Jan 8 at 17:37
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    $\begingroup$ No, you're writing an equation of $1$-forms, and they are denoted only with $d$. $\endgroup$ – Ted Shifrin Jan 8 at 17:38
  • $\begingroup$ @TedShifrin See my edit, in my opinion this is irrelevant for the question. $\endgroup$ – ConvexHull Jan 8 at 17:49

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