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It appears to me that

$\int \frac{{dx}^2}{dy}$ can be rewritten as

$\int \frac{dx}{dy}\cdot{}dx$ which in turn can be rewritten as

$\int f^{-1}{^\prime}(y)\cdot{}dx$. (it is assumed that $y=f(x)$ although y may not be a function of x; multiple solutions for y may exist for a given value of x if y is, say, $\pm\sqrt{x}$).

Last time I checked, $\frac{dy}{dx}$ is a psuedo fraction, and taking the integral of the derivative of the inverse function of y as a function of x with argument y is possible. But I'm new to integral calculus so I could be mistaken :)

So for a more tangible example, assume $y=x^2$

then $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}=\frac{1}{2x}=\frac{1}{2\sqrt{y}}=\frac{d}{dy}\left(\sqrt{y}\right)=\frac{d}{dy}\left(x\right)$

so $\int \frac{dx}{dy}\cdot{}dx = \int\frac{1}{2x}\cdot{}dx=\frac{ln(x)}{2}+C$

What my question is, is can you just do that? rewrite $dx\cdot{}dx$ as ${dx}^2$? I'm fairly confident the rest of my explanation is logically sound. If you feel you know the answer to my original question, feel free to answer it! If you think I am mistaken somewhere, please explain precisely where and don't just accuse me of nonsensically tinkering with mathematical syntax that "doesn't work that way" without explaining where exactly I'm mistaken.

cmarangu
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  • @OP, how would this work in your opinion with the fundamental theorem of calculus? $$\frac{d}{dx}\int_{a}^{x}f(t)~dt = f(x)$$ – Vaas Oct 13 '19 at 02:34
  • @Vaas since we are sort of adding another variable into the proof, give me a minute to consider this – cmarangu Oct 13 '19 at 02:44
  • Have you seen [this question](https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio/21209#21209)? Have you gone through the definition of a differential operator in your studies? – Axion004 Oct 13 '19 at 02:50
  • @Axion004 no I have not gone through the definition of a differential operator thoroughly, I just assumed they were like a sort of pseudo-infinitesimal which, in order to simplify, you have to take a limit of a variable or something – cmarangu Oct 13 '19 at 02:52
  • Firstly, $\int\frac{dx}{dy}\cdot dx=G(x)$ because I am not sure how to simplify this without defining y and x. So if we just mix it in with the equation from the FTC we get $\frac{d}{dt}\left(\int_0^t\frac{dx}{dy}\cdot{}dx\right)=G^{\prime}(t)=g(t)$. QED that they can coexist. – cmarangu Oct 13 '19 at 02:58

3 Answers3

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Without any explanation to the contrary, the conventional meaning of "$\frac{\mathrm{d}x^2}{\mathrm{d}y}$" is $\frac{\mathrm{d}}{\mathrm{d}y} x^2$. If $x$ is independent of $y$, this is zero. If $x$ depends on $y$, then $$\frac{\mathrm{d}}{\mathrm{d}y} \left( x(y) \right)^2 = 2x(y)\frac{\mathrm{d}x}{\mathrm{d}y} \text{,} $$ by the chain rule. Applying this to your integral example seems to lead us astray very quickly.

You seem to want to interpret this differently. My objection to your interpretation is: as you seem to be using it, "$\frac{\mathrm{d}x}{\mathrm{d}y}$" is a single indivisible object representing a limit operation used to find the rate of change of $x$ with respect to variation in $y$. It doesn't have a numerator or a denominator any more than a minus sign, ${}-{}$, does. Consequently, the interpretation you are proposing for "$\frac{\mathrm{d}x^2}{\mathrm{d}y}$" is gibberish. When you use $\frac{\mathrm{d}x}{\mathrm{d}y}$ to represent the derivative of $x$ with respect to $y$, the way to write "$\frac{\mathrm{d}x}{\mathrm{d}y} \cdot \mathrm{d}x$" is exactly that way.

Eric Towers
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  • Fair enough. But since it is a pseudo fraction and dx can be thought of as a sort of infinitessmal variable which is porportional to dy and f(x), I still argue that if superscript was not used to denote second, third, nth derivatives like $\frac{dx^2}{dy^2}=\frac{d}{dy}\left(\frac{dx}{dy}\right)$, then it would be valid to say that one could just multiply the numerator of the pseudo-fraction by the dx variable/constant/whatever it is not yet in the numerator to get my original $\frac{dx^2}{dy}$ pseudo-infinitessimal variable/ratio – cmarangu Oct 13 '19 at 02:17
  • @cmarangu : The second derivative is $\frac{\mathrm{d}^2 x}{\mathrm{d}y^2} = \frac{\mathrm{d}}{\mathrm{d}y} \frac{\mathrm{d}}{\mathrm{d}y} x$. The superscript you used in your question is a power of $x$, it is unrelated to higher derivatives. – Eric Towers Oct 13 '19 at 02:22
  • @cmarangu : Also, I have looked at my usual sources for mathematical nomenclature, and I do not find a definition of pseudofraction compatible with the derivative. The derivative is not any kind of fraction; it is a limit. – Eric Towers Oct 13 '19 at 02:25
  • I edited the Latex code so that if someone was to read it, they would not struggle confusing my "(dx)²" with "d(x²)" – cmarangu Oct 13 '19 at 02:41
  • @cmarangu : This produces no change in the visible output, so a normal reader cannot see any difference. Generally, no one reads the source $\LaTeX$... – Eric Towers Oct 13 '19 at 15:19
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Your interpretation of $dx^2$ as $(dx)(dx)$ may get confused with $ d(x^2)$.

Otherwise your thought process is straightforward and worths further investigation.

You example makes perfect sense.

Mohammad Riazi-Kermani
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There has been a long history of deriding the treatment of $\frac{dy}{dx}$ as a fraction. In the 1800s, they did not have a rigorous methodology for treating infinitesimals, and George Berkeley's attack on them (calling them "ghosts of departed quantities") prevented their common usage. However, in the 1960s, non-standard analysis arose, which gives a rigorous methodology for the treatment of infinitesimals, and developed the hyperreal number system.

These systems allow for $\frac{dy}{dx}$ to represent an actual fraction of infinitesimals. Thus, your system above is perfectly workable. Now, keep in mind that since infinitesimals have not been treated as fractions for a while, some amount of baggage is in place which you may have to carve away yourself. For one example of this, see Extending the Algebraic Manipulability of Differentials. For one on partial differentials, see Exploring Alternate Notations for Partial Differentials. To see how these work together to make for a more straightforward version of calculus, see Simplifying and Refactoring Introductory Calculus.

In short, feel free to use differentials as algebraic units, but remember to double-check what you are doing, as someone may have assumed that you won't do that in their notation. In your case, to avoid confusion, you might want to write $dx^2$ as $(dx)^2$ to be more clear.

As an example of the issues you may run into (from the first paper above), in order to use higher order derivatives algebraically, you have to write them differently. The second derivate of $y$ with respect to $x$ when written in an algebraically manipulable way is: $$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$

The derivation for that is given in the paper, but in short it is a simple application of the quotient rule to the first derivative.

johnnyb
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    Thank you for those articles/websites, but what I do not understand is this "algebraically manipulable" form of the second derivative of y with respect to x. I cannot diffrentiate from the second-derivative superscript 2's from the superscript 2's meaning an infinitessimal variable is being squared, from the superscript 2's meaning a function is being squared or called upon itself – cmarangu Oct 17 '19 at 14:42
  • @cmarangu So, some of this is shorthand. So, to begin with, $dx$ is shorthand for $d(x)$ - the differential operator applied to $x$. Superscripting an operator is unfortunately ambiguous - it can either mean squaring the result of the operator (as is often the case with, say, $\sin^2(x)$), or it can mean performing the operation twice, which is the case with $d^2x$. $d^2x = d(d(x))$. Also, to be more precise, $dx^2 = (d(x))^2$. These are both elaborated (with additional detail) in the links above, especially the first and third link. – johnnyb Oct 17 '19 at 19:50
  • Incidentally, I only recently came to realize that d(y) = dy and that is how we get d/dx (y) = dy/dx myself. Ok cool. Also I do not understand why, in the part after the − sign, you don't just do dx³ instead of dxdx² – cmarangu Oct 19 '19 at 21:24
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    @cmarangu - it's perfectly acceptable to coalesce the right-hand side to $\frac{dy\,d^2x}{dx^3}$, but separating it allows you to better see various ratios. For instance, you can see the first derivative show up, and then you can see that it is multiplied by the second differential of x divided by the square of the first differential. Anyway, I think those ideas will be important in the long-run, so that's why it is written the way that it is. – johnnyb Oct 20 '19 at 00:32