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:

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

  2. What conditions on $f$, $g$ and $h$ are needed to guarantee that

    $$\frac{df}{dg}\frac{dg}{dh} = \frac{df}{dh}?$$

  3. 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

$$\frac{df}{dg} \frac{dg}{df} = 1?$$

  • $\begingroup$ Hint: If you add in $\frac{u}{u}$ you'll get 1/g'*f' $\endgroup$ – Alex Youcis Aug 13 '13 at 6:58
  • $\begingroup$ Am I just too naïve with my limits, or do we have $$ \frac{df}{dg}(x) = \lim_{u\to 0}\frac{f(x + u) - f(x)}{g(x + u) - g(x)} = \lim_{u\to 0}\frac{\frac{f(x + u) - f(x)}{u}}{\frac{g(x + u) - g(x)}{u}} = \frac{f'(x)}{g'(x)} $$ $\endgroup$ – Arthur Aug 13 '13 at 7:48
  • $\begingroup$ Have you heard of Stieltjes Integral? I think they have some in common. By nowhere vanishing do you mean g or dg? $\endgroup$ – eccstartup Aug 13 '13 at 8:52
  • $\begingroup$ I don't see the last equality holds. @Arthur $\endgroup$ – eccstartup Aug 13 '13 at 8:53
  • 1
    $\begingroup$ @eccstartup $\lim a$ and $\lim b$ in this case would be $f'(x)$ and $g'(x)$ by the definition of the derivative. If they exist or not is a different story. $\endgroup$ – Arthur Aug 13 '13 at 17:10

Regarding Questions 1 and 3, it's not crucial that $g$ be non-vanishing in $X$; after all, $x$ (i.e., the identity function) vanishes. However, it is important that $g$ be locally one-to-one (and therefore strictly monotone if $g$ is continuous and the domain $X$ is an interval) in order for $df/dg$ to be defined throughout $X$.

The answer to Question 3 is "no": Take $f(x) = 1 + x^3$ and $g(x) = 1 + x^5$ on $X = (-1, 1)$. ($dg/df(0) = 0$ and $df/dg(0)$ is undefined. The added "$1$"s guarantee $f$ and $g$ are non-vanishing, but have no effect on the difference quotient for $df/dg$.)

Here's are some (variously far-flung) musings as to why this definition of $df/dg$ is not widespread.

A derivative $df/dx$ may be viewed as measuring the rate of change of $y = f(x)$ against a uniform standard rate, namely $1 = dx/dx$.

To compare the relative rates of change of $y = f(x)$ and $z = g(x)$ (when this makes sense), one is naturally led to compute a ratio of derivatives: As Arthur notes, if $f'(a) = df/dx(a)$ and $g'(a) = dg/dx(a)$ exist for some $a$, and if $g'(a) \neq 0$, then $df/dg(a)$ exists and is equal to the ratio $f'(a)/g'(a)$. (These observations permeate the discussion so far; just spelling them out explicitly.)

One philosophical point in the two preceding paragraphs is the analogy to the problem of converting data (say, image or audio files) between multiple formats: It's easiest to create a single master format $M$, and to convert format $A$ to format $B$ by converting $A$ to $M$, then $M$ to $B$.

In this analogy, $y = f(x)$ is format $A$, $z = g(x)$ is format $B$, and $x$ is the master format $M$; "converting" $A$ to $B$ means "taking the rate of change", and breaking up the conversion corresponds to the formula $df/dg = (df/dx)/(dg/dx)$.

To complete this train of thought, "$df/dg$" can be accommodated (modulo details, see below) within the framework of differential calculus.

As to precise hypotheses: If the domain $X$ is an interval and if $f$, $g$, and $h$ are differentiable with non-vanishing derivatives, then the answer to all three questions is "yes". However, these are restrictive hypotheses, and they appear to be "(close to) minimal" if one wants all three questions to have an affirmative answer.

Of course, these hypotheses are not necessary if one merely wants $df/dg$ to be defined, but if $h$ is differentiable and $f = h\circ g$, then existence of $df/dg$ implies relatively little about $g$. In an extreme case, $h(x) = x$, $g$ is injective (but otherwise as pathological as you like), and the difference quotient for $df/dg$ is identically equal to $1$.

That is, the proposed definition of $df/dg$ "works entirely as expected" only under restrictive hypotheses (that are already handled well by differential calculus), but in general is too lax to put any substantial restrictions on $g$.


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.

  • $\begingroup$ It doesn't matter that $f$ and $g$ tends to $x$ in the same pace. $\frac{df}{dg}(x)$ might not be the definition of $\frac{f'(x)}{g'(x)}$, but as long as $f'(x)$ and $g'(x)$ exist and $g'(x)\neq 0$, the limits are the same, by the reasoning in my comment above. $\endgroup$ – Arthur Aug 13 '13 at 10:55

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