Differentiating one function with respect to another. Given an open set $X \subseteq \mathbb{R}$ and total functions $f,g : X \rightarrow \mathbb{R},$ we can define a partial function $$\frac{df}{dg} : X \rightarrow \mathbb{R}$$
by asserting that
$$\frac{df}{dg}(x)=\lim_{u \rightarrow 0} \frac{f(x+u)-f(x)}{g(x+u) - g(x)}$$
The domain of definition of  $\dfrac{df}{dg}$ can be taken as the set of all $x \in X$ such that the above limit exists.
Remark. If we're writing $x$ for the identity function on $X$, then the expression $\frac{df}{dx}$ means what you'd expect it to mean, namely $f'$.
BUT, I have not seen this definition anywhere.
Is there something wrong with it?
And if not, where can I learn more? I would like to read an article that systematically builds up the theory of this operator.
A few specific questions:

*

*Suppose $f$ and $g$ are differentiable and that $g$ is nowhere vanishing. Is $\frac{df}{dg}$ necessarily a total function?


*What conditions on $f$, $g$ and $h$ are needed to guarantee that
$$\frac{df}{dg}\frac{dg}{dh} = \frac{df}{dh}?$$


*Suppose $f$ and $g$ are differentiable and nowhere vanishing. Is it necessarily the case that $\frac{df}{dg}$ and $\frac{dg}{df}$ are total functions satisfying
$$\frac{df}{dg} \frac{dg}{df} = 1?$$

In the common sense of chain rule, we need $f(g)$ and $g(h)$ differentiable to have $\frac{df}{dg}\frac{dg}{dh} = \frac{df}{dh}$ holds (denominator non-vanishing required). In your definition, $f$ and $g$ tends to $x$ in the same pace $u$, which is (probably) not the definition of $\frac{f'(x)}{g'(x)}$. It does reduce to $f'(x)$ when $g(x)=x$, but I don't see any use of this definition.
Well, I think it can be viewed as one kind of derivative which extends the common derivative. Sometimes it can be useful when you seek a structure while common sense of derivative don't have. Maybe you can find some more interesting results to make it useful.
