Why is this a fake proof?

Question

I am aware of the "definition" of the total differential as follows:

$$\mathrm{d}f = \frac{\partial f}{\partial x} \mathrm{d}x + \frac{\partial f}{\partial y} \mathrm{d} y.$$

Now, assume we wished to show that:

$$\frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{\partial f / \partial x}{\partial f / \partial y}$$

for a level curve of $f$. Now, I thought one could accomplish this as follows: For a level curve, $f(x, y) = c$, we have that $\mathrm{d}f = 0$, so:

$$\frac{\partial f}{\partial x} \mathrm{d}x + \frac{\partial f}{\partial y} \mathrm{d} y = 0$$

and so therefore:

$$\frac{\mathrm{d} y}{\mathrm{d} x} = - \frac{\partial f / \partial x}{\partial f / \partial y}$$

by simple manipulation. I was told that this isn't rigorous, and furthermore, that's it entirely incorrect in nature. Now, I understand that it's nonrigorous, because it involves loosely using $dx$ and $dy$, but I'm sure there is some way to make the above treatment of differentials more rigorous, but it probably involves concepts that I haven't learned yet. What is the logical flaw in this "proof"?

Answer 1

$\mathrm dx $ and $\mathrm dy$ can't be taken as meaningless symbols. If they were, how could you talk about what it means to divide them (or even add them)? How could you talk about $\mathrm df$ as a linear combination of them, unless that also makes it a meaningless symbol?

No, these are well-defined quantities called differential forms.

A differential $k$-form is a function of the coordinate position as well as a function of $k$ vectors. $\mathrm dx$ and $\mathrm dy$ are differential 1-forms, and the equation you were given for the total differential can be seen as defining $\mathrm df$ as a differential 1-form in terms of $\mathrm dx$ and $\mathrm dy$:

$$\begin{align*}\mathrm df&: \mathbb R^2 \times \mathbb R^2 \to \mathbb R\\ \mathrm df(r,a) &= \partial_1 f(r) \mathrm dx(r, a) + \partial_2 f(r) \mathrm dy(r,a)\end{align*}$$

This is a very verbose and explicit way of writing the total differential. I'm being extremely, extremely pedantic here: I'm not even calling $\partial_1 f$ by $\partial f/\partial x$ because, from a strict mathematics perspective, $f$ is a function of a vector argument, and the components of that argument (the coordinates) need not be called $x,y$.

When we consider a level curve of $f$, that means we have some curve $C: \mathbb R \to \mathbb R^2$ such that $(f \circ C)(t) = K$ for some constant $K$. Taking a derivative of this function yields, using the chain rule,

$$\begin{align*}0 &= (f \circ C)'(t) \\ &= \mathrm df(C(t), C'(t)) \\ &= [(\partial_1 f) \circ C](t) \mathrm dx(C(t), C'(t)) + [(\partial_2 f) \circ C](t) \mathrm dy(C(t), C'(t))\end{align*}$$

Again, being painfully explicit here. Most people would not even bother writing the function $C$ in here, and the derivative $C'$ would be considered implied.

Now at this point, you can write

$$\frac{\mathrm dy(C(t), C'(t))}{\mathrm dx(C(t), C'(t))} = - \frac{(\partial_1 f)\circ C(t)}{(\partial_2 f) \circ C(t)}$$

That, however, is far from considering $y$ a function of $x$ and taking a derivative.

Now, what some people might do is define the level curve such that $C$ takes one of the coordinates (like $x$) and spits out the coordinate pair $(x,y)$ that corresponds to a point on the curve. When this is done, $C(x) = (x, Y(x))$ and $C'(x) = (1, Y'(x))$ for some function $Y$. Moreover, $\mathrm dx$ and $\mathrm dy$ don't actually depend on position at all: they only look at the second argument.

When this is done, the resulting equation looks like

$$Y'(x) = -\frac{\partial _2 f(x,Y(x))}{\partial _1 f(x,Y(x))}$$

So while the argument put forth is very powerful and very suggestive, and it gets the gist of things right, the bookkeeping required "under the hood" of this argument may require some additional thought.