Can we differentiate a function w.r.t itself?

Question

A question came to my mind which is: is it possible to differentiate a function with respect to itself?

I tried to think of how it will work and came to this.

I begin with \begin{gather} f(x)=f(x). \end{gather}

It's kind of silly but bear with me.

Then I've applied the definition

\begin{gather}\lim_{f(x) \to f(a)} \frac{f(x)-f(a)}{f(x)-f(a)}=\lim_{f(x) \to f(a)} \frac{f(x)-f(a)}{f(x)-f(a)}\end{gather}

the L.H.S will be equal to $1$. Then,

\begin{gather}1=\lim_{f(x) \to f(a)} \frac{f(x)-f(a)}{f(x)-f(a)}.\end{gather}

The trick I've made is: as ${f(x) \to f(a)} \equiv x\to a$.

Then I rewrote the expression \begin{gather} \lim_{x \to a} \frac{f(x)-f(a)}{f(x)-(a)} =\lim_{f(x) \to f(a)} \frac{f(x)-f(a)}{f(x)-f(a)},\end{gather}

I multiplied by $x-a$ up and down,

\begin{gather} \lim_{x \to a} \frac{f(x)-f(a)}{x-a}.\frac{x-a}{f(x)-f(a)},\end{gather}

and I end up with this \begin{gather} 1=f'(a) \frac{dx}{df(x)}.\end{gather}

My question is: is it meaningful to have $\frac{dx}{df(x)}$? The independent variable is $x$, so how could we say that we measure the change of $x$ w.r.t. $f(x)$?

My second question: is it possible to apply the chain rule from the beginning? because I know that we can apply chain rule for a function of some variable, like $f(x)$ for example. But could we apply the chain rule for an independent variable of this function, like $x$ is the independent variable of $f(x)$. Could we say from the beginning $1=f'(x)\frac{dx}{d(f(x))}$, and why?

Answer 1

$\frac{dx}{d(f(x))}$ is completely valid if $x$ can be expressed as a function of $f(x)$, i.e., $x=g(f(x))$.

This function $g$ is the inverse of the function $f$, i.e. $f^{-1}$, hence the function $f$ must be invertible.

$g(x)$ must also be a differentiable function with respect to $x$.

Answer 2

As a general rule, the collection of symbols $$ \frac{\mathrm{d}f(x)}{\mathrm{d}x}$$ is not a fraction—it is not the ratio of two real numbers, for example. There are many different ways of explaining precisely what this notation means (e.g. as a fraction of hyperreal numbers, in the sense of Abraham's nonstandard analysis; or as the result of an operator acting on the function $f$; or a bunch of other things), and there are many abuses of this notation which give good heuristics for true results (such as the chain rule). However, at the level of introductory calculus, this object should be treated as a single symbol, which represents the evaluation of the derivative function at a specific point.

What this means is that the step in the question $$ \lim_{x\to a} \frac{x-a}{f(x) - f(a)} = \frac{\mathrm{d}x}{\mathrm{d}f(x)} $$ is, essentially, nonsense. It doesn't really mean anything in terms of the way in which these symbols are typically defined.

That being said, it is possible to give a reasonable meaning to this notation via the chain rule: if $f$ and $g$ are both differentiable functions, then, via a very slight abuse of notation, $$ \frac{\mathrm{d}(f\circ g)(x)}{\mathrm{d}x} = \frac{\mathrm{d} f( g(x) )}{\mathrm{d}g(x)} \cdot \frac{\mathrm{d}g(x)}{\mathrm{d}x}. $$ Thus, in the context of the chain rule, it can make sense to write something like $$ \frac{\mathrm{d}x}{\mathrm{d}f(x)}. $$ However, note that I am thinking of this in the same way that I think about the notation used for the chain rule: to write $ {\mathrm{d}g}/{\mathrm{d}f}$ (where $f$ and $g$ are functions) is to implicitly assume that $g$ is a function of $f$.

Therefore, if I write $$\frac{\mathrm{d}x}{\mathrm{d}f(x)}, $$ I am implicitly assuming that $x$ is a function of $f(x)$, i.e. there exists some differentiable function $g$ so that $x = g(f(x))$. But the functional equation $x = g(f(x))$ has a solution! By definition, $$ x = g(f(x)) \implies g = f^{-1}. $$ In other words, if the notation $\mathrm{d}x/\mathrm{d}f(x)$ is going to have any meaning, then it almost certainly must mean the same thing as $$ \frac{\mathrm{d}f^{-1}(f(x))}{\mathrm{d}f(x)}. $$ But this notation has a very reasonable meaning in terms of the chain rule: $$ \frac{\mathrm{d}(f^{-1}\circ f)(x)}{\mathrm{d}x} = \underbrace{\frac{\mathrm{d}f^{-1}(f(x))}{\mathrm{d}f(x)}}_{=\mathrm{d}x/\mathrm{d}f(x)} \cdot \frac{\mathrm{d}f(x)}{\mathrm{d}x} = (f^{-1})'(f(x)) \cdot f'(x). $$ As such, it would appear that if the notation $\mathrm{d}x/\mathrm{d}f(x)$ is sensible notation, then the only thing it can reasonably mean is "the derivative of the inverse of $f$, with respect to its argument, $f(x)$".

Finally, it is argued in the original question that it should be the case that $$ 1 = \frac{\mathrm{d}x}{\mathrm{d}f(x)} \cdot \frac{\mathrm{d}f(x)}{\mathrm{d}x} =(f^{-1})'(f(x)) \cdot f'(x) \implies f'(x) = \frac{1}{(f^{-1})'(f(x))}, $$ which is basically a restatement of the Inverse Function Theorem. As such, we have stumbled into a really good heuristic argument for the Inverse Function Theorem. This isn't quite a proof (there are some assumptions and hypotheses which would need to be cleaned up a bit, and the argument doesn't completely hold water; but these problems can be fixed), but the overall argument is fairly convincing.