Rate of change of a multivariable function with respect to a change in another multivariable function

Question

Is there a general way to proceed with finding the derivative of a multivariable function $g:\mathbb{R^2} \to \mathbb{R}$ with respect to another multivariable function $f:\mathbb{R^2} \to \mathbb{R}$ where both $f$ and $g$ have as arguments $x \in \mathbb{R}$ and $y \in \mathbb{R}$ ? Does the following notation even make sense: $$\frac{d g(x,y)}{d f(x,y)}$$ I have looked at directional derivatives but they do not seem to fit the bill. Similar questions are here, here, here and here. But some of the suggested solutions are specific cases.

Answer 1

Please do not ever use such notation (as Ted Shifrin mentions in the comment of your third link). Here are two concepts which we can speak of from differential geometry:

• Given a function $f$ (from a manifold into reals, for example, $f:\Bbb{R}^2\to\Bbb{R}$), we can speak of its exterior derivative $df$ (or if your domain is a vector space like $\Bbb{R}^n$, you can speak of its Frechet derivative). See How does the idea of a differential $dx$ work if derivatives are not fractions for more details.
• Next, and this is essentially what is mentioned by Qiaochu Yuan in your third link, given a coordinate system $(U,\varphi=(\xi_{\varphi}^1,\dots, \xi_{\varphi}^n))$ on a set $U$ (i.e a nice way of mapping points in your set to a collection of numbers; see this answer of mine for a much more complete description), there is a collection of vector fields $\{\mathbf{v}_{\varphi,1},\dots, \mathbf{v}_{\varphi,n}\}$ which are naturally associated to that coordinate system, i.e for each point $p\in U$ in your set, you have a basis of vectors $\{\mathbf{v}_{\varphi,1}(p),\dots, \mathbf{v}_{\varphi,n}(p)\}$. Given these vectors, we can plug them into $df$ to get the number $df_p(\mathbf{v}_{\varphi,i}(p))$. This number tells us how much the function is changing, instantaneously (more accurately, to first order) at the point $p$, along the direction $\mathbf{v}_{\varphi,i}(p)$. Now, the notation $\mathbf{v}_{\varphi,i}$ is not standard. The standard notation is $\mathbf{v}_{\varphi,i}:=\frac{\partial}{\partial \xi_{\varphi}^i}$ to denote the $i^{th}$ vector field induced by the coordinate system $(U,\varphi)$, and instead of $df_p(\mathbf{v}_{\varphi,i}(p))$, we write $\frac{\partial f}{\partial \xi_{\varphi}^i}(p)$. Often, we get lazy and drop the subscript $\varphi$, and write simply $\frac{\partial f}{\partial \xi^i}(p)$, but it should be kept in mind that this number depends on the other coordinate functions $\xi^1,\dots, \xi^{i-1},\xi^{i+1},\dots, \xi^n$ (again something which is mentioned in your third link).

Now, given two (sufficiently well-behaved) functions $f$ and $g$, we can consider their exterior derivatives $df$ and $dg$. We can now fix a point $p$, and ask whether or not $df(p)$ and $dg(p)$ are proportional (if we're only considering functions of one variable, i.e $f,g:\Bbb{R}\to\Bbb{R}$, then they will always be propertional). In higher dimensions, this need not be the case. If they so happen to be proportional, you can could define the "quotient" modulo some techinicalities with division. But, just because you can, doesn't always mean you should, and here you really shouldn't because in linear algebra we don't work with division of (co)vectors. Another reason we don't bother defining this quotient is that it doesn't even make sense generally! Like I said above, in higher dimensions, it need not be true that the linear maps $df_p$ and $dg_p$ are proportional.

The bottom line is that we rarely ever really mean changing one function with respect to another, even though that is often what people say. Rather what we actually mean and care about is how a function changes along various directions (directional derivatives), and as a whole (Frechet derivatives, exterior derivatives of functions). A very common special case of directional derivatives is the (tangential) directions which are induced by a given coordinate system. To make sense of all this more formally and generally though, you'll first need a good grasp of linear algebra, because differential calculus is literally the study of "local linear approximations", so understanding exactly what happens at the linear level is a prerequisite.