The Nature of Differentials and Infinitesimals

Question

I have been wondering for some time what the limits of Leibniz notation is, and what exactly its meaning is. I learned limits and later learned (to some extent) infinitesimals, but there are some oddities which have me befuzzled. The one person I know who could answer the question gave me a reference so dense I couldn't make heads or tails of it.

In any case, let's say you have a function $y = f(x)$. Now, the derivative is $\frac{dy}{dx} = f'(x)$ and the second derivative is $\frac{d^2y}{dx^2} = f''(x)$. Anyway, if you play around with these a bit, you can see that $\frac{dx}{dx} = 1$, which means that $x$ always changes in unity with itself. However, a very odd result happens if you look at the second derivative. Since $\frac{dx}{dx} = 1$, and 1 is a constant, that means that the second derivative, $\frac{d^2x}{dx^2} = 0$, which means that x never has any acceleration with respect to itself.

However, algebraically, what this seems to mean to me is that $d^2x$ is always zero, but this is obviously not the case, as it could be put in ratio with $dy^2$ to produce a real-valued function. However, this seems to be at odds with an infinitesimal definition of $d^2x$ (or any other definition I have seen). It seems to imply that that $dx$ is more of a relational quantity than an infinitesimal or even a limit.

I did not know if anyone had any specific knowledge about this, or knew of any books that dealt with this topic. I have a hard time finding any at all that approach this subject.

On a side note (but related), I would also be interested in any books which discussed any possible meaning of quantities like $\frac{d^2y}{d^2x}$ (note that this is different from the Leibniz second derivative which is $\frac{d^2y}{dx^2}$). Anyway, if anyone has ideas or references, I would love to investigate this topic further.

Answer 1

You seem to be adding some context to the notation that was never meant to be there.

$\frac{\mathrm{d}}{\mathrm{d}x}$ is just an operator.$\frac{\mathrm{d}^2}{\mathrm{d}x^2}$ is just the notation used when operating twice.

In the language of infinitesimals, you can consider

$\frac{\mathrm{d}y}{\mathrm{d}x}$ to be the infinitesimal change in $y$ and how it compares to $x$ (in a ratio).

I took a course in my second year of University which addressed the differential, differential forms (one-forms, two-forms, etc) and wedge products. It's a slippery slope, but you might want to look at a textbook from a Calculus III course. I can't share mine, as my Professor was the author (I'm not sure if I'm able to distribute this book).

Answer 2

The algebraic definition of the second derivative of $y=f(x)$ is $$ f''(x)=\frac{d\left[\frac{dy}{dx}\right]}{dx} $$ This can be expanded using the quotient rule $$ f''(x)=\frac{d^2y}{dx^2}-\frac{dy\ d^2x}{dx^3} $$ Furthermore, if you desire to evaluate $d^2y/d^2x$ this can be found by taking algebraic differentials $$ dy=f'(x)dx $$ $$ d^2y=f''(x)dx^2+f'(x)d^2x $$ And finally $$ \frac{d^2y}{d^2x}=f''(x)\frac{dx^2}{d^2x}+f'(x) $$ Unfortunately, it is not possible to place a value on this expression because the quantity $d^2x/dx^2$ cannot be evaluated. In order to obtain a usefull result, algebraic manipulation of the derivitive quantities must be done to cancel out all of the differential terms ($dx$, $dy$, $d^2x$, etc.), only then the expression can be evaluated.

Answer 3

$\frac{dx}{dx} = 1$ because the infinitesimal $dx$ changes at the same rate as $dx$.

$\frac{d}{dx}\frac{dx}{dx} = 0$ because $\frac{dx}{dx}$ changes infinitely slower than $dx$. One way to represent this is $\frac{d}{dx}\frac{dx}{dx} = lim_{n \rightarrow \infty} \frac{(\frac{1}{n})}{1} = 0$

The same holds true for $\frac{d^2 x}{dx^2}$. It's not that $d^2 x = 0$, rather, it is that $dx^2$ changes infinitely faster than $d^2 x$. Both are infinitesimals, so on their own they both approach zero. But the ratio approaches zero.

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Some good reading is found on another page: Is There a Difference Between $d^2x$ and $(dx)^2$?

Answer 4

The key term here is second difference. If $y=f(x)$ then the meaning of the symbol $d^2y$ is the infinitesimal version of the second difference $\Delta^2y=f(x)-2f(x+\Delta x)+f(x+2\Delta x)$. If $f\in C^2$ then the second derivative $f''$ can be calculated in terms of the second difference by taking the limit of $\frac{\Delta^2 y}{\Delta x^2}$. This explains Leibniz's notation for second derivative that the OP mentioned.

By far the clearest discussion of this is in Keisler's textbook Elementary Calculus, page 94.

Answer 5

After a lot of searching, I realized that the answer to this question depends on which variable is the independent variable. $d^2u$ reduces to $0$ when $u$ is the independent variable, but, otherwise, it should stay.

The problem is that the second derivative as it is generally taught is actually a special case. The real notation for the second derivative of $x$ with respect to $y$ should be $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. If (and only if!) $x$ is the independent variable, then $d^2x$ reduces to $0$ and the second term goes away. However, if you leave it in the expanded form, you can do a lot of cool tricks with it, like algebraically manipulate it into the second derivative of $x$ with respect to $y$.

Anyway, I wrote a paper on my investigations into this, and you can see it here:

https://arxiv.org/abs/1801.09553

Answer 6

$d^2 x$ is not "always" zero. It depends on the context. With respect to itself, sure $\frac{d^2 x}{dx^2} = \frac{d}{dx}\frac{dx}{dx} = 0$ but suppose with respect to $y$ it isn't 0. Then $\frac{d^2 x}{dy^2} = \frac{d}{dy}\frac{dx}{dy} \ne 0$ then how could $d^2x = 0$ ?