Is $\int \frac{{dx}^2}{dy}$ valid?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.