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The operator $\frac{d}{dx}$ is common in calculus to denote a derivative. However, this also begs the question, what is the operator $\frac{dx}{d}$? Is this operator used commonly? If so, what is it called/what does it do?

I have played aroud with it before, and found a natural way to define it seems to be that $$\frac{dx}{d}\frac{1}{f(x)} := \frac{dx}{df} = \frac{1}{(\frac{df}{dx})}$$

I found also in my own playing around that this could define an odd thing when applying the operator twice:

$$\frac{dx}{d}(\frac{dx}{df}) = \frac{dx}{d\frac{df}{dx}} = \frac{1}{\frac{d\frac{df}{dx}}{dx}} = \frac{1}{\frac{d^2f}{dx^2}}$$

Which would seem to imply a nice notation definition:

$$\frac{dx}{d}(\frac{dx}{df}) := \frac{dx^2}{d^2f}$$

All this is purely my own speculation/invention, of course. I've never heard of any operation like this, and can't find it on the internet, because I don't have a name for it and can't find the notation anywhere. Is this operation already well-defined?

Michael Hardy
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MathTrain
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    $\frac{d}{dx}[f(x)]$ is shorthand for the following definition: $\frac{d}{dx}[f(x)]:=\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}$. It is not technically a fraction, nor should one expect it to behave like one. – JMoravitz Aug 01 '16 at 20:01
  • @JMoravitz I do not expect it to behave as a fraction always, though it often does in many cases, which is why it's a useful notation. I'm simply asking if there is a separate notation which uses this. So far it seems this is not used, though I couldn't necessarily see a problem with defining it as I've tried to above? Though I'm not sure – MathTrain Aug 01 '16 at 20:07
  • Just have to comment because of a pet peeve of mine: “begs the question” refers to a logical fallacy that’s basically circular reasoning. It is not synonymous with “prompts the question.” And now, back to answers to the question that was prompted by $\frac d{dx}$. – amd Aug 02 '16 at 06:52
  • @amd that's nice, I'll be more careful next time, but i'm just wondering if you happened to learn about that from the Crash Course Philosophy video that came out a day before you posted this. Could be a coincidence but I'm curious – MathTrain Aug 03 '16 at 04:50
  • Coincidence. This is something that’s made my teeth itch for many years. – amd Aug 04 '16 at 06:22

2 Answers2

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One problem is that viewing $\text{“ } \dfrac {dx} d\text{ ''}$ as an "operator" is that it should potentially be able to operate on something, thus: $$\text{“ } \frac {dx} d f \text{ ''}. \tag 1$$

That makes sense with $\dfrac d{dx}$ because $\dfrac d {dx} f$ means $\dfrac{df}{dx}$. But nothing analogous can be done with the expression in $(1)$.

Michael Hardy
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    For the record, $\frac{dx}{d} \frac{1}{f}$ doesn't mean anything either :) – Clement C. Aug 01 '16 at 20:05
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    ok, Let's see if we can demolish the following idea: $$ \frac{dx} d, f = \frac{dx} d, \frac 1 {1/f} = \frac{dx}{d(1/f)} = -f^2 \cdot \frac{dx}{df} $$ – Michael Hardy Aug 01 '16 at 20:22
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    Actually in MathTrain's definition, $\frac{dx}{d}$ does describe a mapping, namely $\mathrm{inv}\circ\frac{d}{dx}\circ\mathrm{inv}$ where $\mathrm{inv}: (x\mapsto f(x))\mapsto(x\mapsto\frac{1}{f(x)})$. And while $\mathrm{inv}$ is not a linear function, applying it twice "cancels" its nonlinearity out (well, except for the factor $0$, but that could be defined by continuity), so $\frac{dx}{d}$ is even linear, hence an operator. I don't think the operator is particularly useful, but then, "being useful" is not part of the definition of an operator. – celtschk Aug 01 '16 at 20:45
  • @MichaelHardy I don't think you can demolish it, I tried it for f(x)=x^3 and it worked perfectly. Seems like it's correct. – MathTrain Aug 01 '16 at 22:20
  • @MathTrain : If you mean the identity $\dfrac{dx}{d(1/f)} = -f^2 \dfrac{dx}{df} $ the the fact that that works is trivial. $\qquad$ – Michael Hardy Aug 02 '16 at 02:59
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First of all, your operator needs to act on some space of functions. The obvious restrictions could be non-vanishing functions with non-vanishing derivative. This immediately rules out linear functions and specifically $f\equiv 0$, so your operator won't act on usual vector space of functions, at least not without suitable modifications (perhaps with something involving extension by continuity, as suggested by celtschk in comments).

For example, if we are talking about $C^1$ functions, "non-vanishing functions with non-vanishing derivative" are positive/negative strictly monotonous functions.

However, this still won't be linear (at least not with standard addition). To see this, I will change notation to $f' = \frac{df}{dx}$ to avoid unnecessary clusters.

So, to definition:

$$\frac{dx}d f = \frac{1}{(\frac 1f)'}= \frac{-f^2}{f'}$$

This is homogeneous:

$$\frac {dx}d (af) = \frac {-a^2f^2}{af'} = a\frac{dx}df,$$

but not additive:

$$\frac{dx}df + \frac{dx}dg = \frac{-f^2}{f'} + \frac{-g^2}{g'} = -\frac{f^2g'+f'g^2}{f'g'}=-\frac{(f^2g'+f'g^2)(\frac 1{f'}+\frac 1{g'})}{f'+g'} = -\frac{f^2 + g^2 + \frac{f^2}{f'}g' + f'\frac{g^2}{g'}}{f'+g'} = \frac{dx}d(f+g) - \frac{\frac{f^2}{f'}g' + f'\frac{g^2}{g'}-2fg }{f'+g'}.$$

Notice that $\frac{f^2}{f'}g' + f'\frac{g^2}{g'}\neq 2fg$ in general, for example, take $f(x) = e^x$ and $g(x) = e^{2x}$.

One could hope then that $\frac{dx}{d(f+g)} = \frac{dx}{df} + \frac{dx}{dg}$ but this is even easier to see to fail.

But, there is even more elementary problem with addition. Sum of functions or their derivatives could vanish at some points. To remedy this, we would have to restrict ourselves even more, for example only to positive functions. But for $\frac{dx}d$ to be well defined then, its derivative should be negative. Thus, we could observe strictly decreasing positive functions and let $\frac{dx}d$ to act on those. It will be (positively) homogeneous, but still not additive.

Ennar
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  • So to add some context, I was working on this in conjunction with another "project" of mine, about which I have made an extensive post before: http://math.stackexchange.com/questions/1785715/finding-properties-of-operation-defined-by-x%E2%8A%95y-frac1-frac1x-frac1y/1785737#1785737. --- It has to do with the operation ⊕, which is defined as x⊕y=1/(1/x+1/y). Using this operation, which is similar to addition in that it is distributive with multiplication, associative, and commutative, we can see that $\frac{dx}{d}f⊕\frac{dx}{d}g = \frac{dx}{d}(f⊕g)$ – MathTrain Aug 01 '16 at 22:43
  • @MichaelHardy Due to your insight that $\frac{dx}{d}f = -\frac{f^2}{f'}$ I have realized that the operator $\frac{dx}{d} $is equivalent to the oplus-derivative which Mike4ty4 invented in his answer to my previous post! Link is http://math.stackexchange.com/questions/1785715/finding-properties-of-operation-defined-by-x%E2%8A%95y-frac1-frac1x-frac1y/1785737#1785737 – MathTrain Aug 02 '16 at 00:25