Equation of Partial derivatives

Question

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 ?

Answer 1

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.

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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, $$ \begin{bmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v} \end{bmatrix} \begin{bmatrix} \dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y} \end{bmatrix} =I\tag4 $$

Answer 2

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.

Answer 3

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