# How to teach Leibniz and Newton's notation

There has been many posts here and in MSE about different notations of differentiation. See for example this, this and this.

However, those questions only deal with the common misunderstanding about Leibniz notation. But none of them really tell how to avoid getting the wrong idea.

So how to teach these different notions about differentials in a way that this misunderstanding doesn't arise? Or should one even care about this because it seems to be a pretty common practice not to (at least in physics)?

• A small historical remark: as far as I know Newton never used the dy/dx notation, nor did he use f(x), nor did he speak of functions or variables. – Michael Bächtold Mar 3 '18 at 13:50
• @MichaelBächtold Thats true. But I didnt claim he did. The idea of the question is to find out best way to teach the two different notation without raising misunderstanding. – Harto Saarinen Mar 3 '18 at 21:09
• sorry I misunderstood then, it seemed like the question was just about the dy/dx notation and differentials. Which misunderstanding about Newtons notations did you have in mind? – Michael Bächtold Mar 4 '18 at 6:20
• @MichaelBächtold Maybe the question isnt written clearly. The idea is to find out the way to teach both of the notations in a way that students learn them in a rigorous way. Newtons notation doesnt cause misunderstandigs too often but its harder for student to use it. Leibniz OTOH causes misunderstandings but is super easy to use. The idea is to find the balance between these and avoid the one misunderstanding. – Harto Saarinen Mar 4 '18 at 14:45
• To clear this up, Newton's derivative notation is to put a dot over the function. It is awkward for compound functions, and I'm not sure anyone uses it these days. The notation $f'(x)$ is Lagrange notation. For more details, en.wikipedia.org/wiki/… – Matthew Daly Dec 12 '19 at 13:43

The reason why so many people get the wrong idea about differentials is that they aren't really taught what the notation means. They are merely taught "this is what the notation is, and please don't ask any deep questions." This is a recipe for misusing the notation. Additionally, some of the standard notations (like for the second derivative) are flat-out wrong, but we will get to that later.

To start out with, you should think of $d()$ as a function. Therefore, $dy$ is actually shorthand for $d(y)$. The differential function can be applied multiple times, such as $d(d(y))$, which is normally written as $d^2y$. So, when you see a notation that says $d^2(y)$ you should think $d(d(y))$ and when you see a notation that says $dx^2$ you should think $(dx)^2$. This alone clears up a LOT of confusions that people have in dealing with the notation.

With this explanation in hand, it becomes obvious and clear why $d^2y$ and $dy$ don't cancel. It's the same reason why you can't cancel with $\sin(\sin(y))$ and $\sin(y)$. In fact, for this reason, when I'm doing technical writing, I usually write the "d" with non-italic text, like $\mathrm{d}y$, so that it is typographically evident that "d" and "y" are playing different roles (I don't do that here, because I have a bunch of LaTeX macros that help me out that I don't have here).

Now, I'm going to go against mainstream wisdom and say that $\frac{dy}{dx}$ absolutely should be treated as a fraction. Doing so simplifies Calculus in a number of ways, which I will show:

• It is more clear to the student what is going on
• It removes the distinction between explicit and implicit differentiation
• It removes the distinction between single-variable and multivariable differentiation

In other words, you get a single, unified, standardized process for all of these types of differentiation.

The way that I teach it is to NEVER take derivatives initially. ONLY take differentials. Then, you can SOLVE for whatever derivative you want. But first, you have to convert your derivative rules into differential rules. So, for a few rules (hopefully you can figure out the rest):

• $d(C) = 0$
• $d(u) = du$
• $d(nu) = n\,du$
• $d(u^n) = n\,u^{n-1}\,du$
• $d(uv) = u\,dv + v\,du$

Now, let's take the equation $z = xy - g$, where all of these letters are actual variables. Let's say we want to find $\frac{dg}{dy}$. Well, the first step NO MATTER WHAT derivative you want to find, is to take the differential of both sides:

$$d(z) = d(xy - g) \\ d(z) = d(xy) - d(g) \\ dz = x\,dy + y\,dx - dg$$

Now, to find the derivative $\frac{dg}{dy}$ we just algebraically manipulate the equation:

$$dz = x\,dy + y\,dx - dg \\ dg = dz - x\,dy - y\,dx \\ \frac{dg}{dy} = \frac{dz}{dy} - x\frac{dy}{dy} - y\frac{dx}{dy} \\ \frac{dg}{dy} = \frac{dz}{dy} - x - y\frac{dx}{dy}$$

Now, the real problem comes with the second derivative. You can't treat that as a fraction. However, this is not because it doesn't work, but because the notation that has been handed down is simply wrong (or, technically, it is only right within a particular context). The second derivative is usually given as:

$$\frac{d^2y}{dx^2}$$

This is only true when x is the independent variable. If you later say that $x$ depends on some other variable, such as $t$, the notation will break down. The full notation for the second derivative (which does allow for treating it as fractions) is:

$$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$

This can be easily shown by just applying the quotient rule to the first derivative and simplifying:

$$d\left(\frac{dy}{dx}\right) = \frac{dx\,d(dy) - dy\,d(dx)}{dx^2} \\ = \frac{dx\,d^2y - dy\,d^2x}{dx^2} \\ = \frac{dx\,d^2y}{dx^2} - \frac{dy\,d^2x}{dx^2} \\ = \frac{d^2y}{dx} - \frac{dy}{dx}\frac{d^2x}{dx} \\$$ That was just the differential--to get the second derivative you also have to divide by $dx$, yielding: $$= \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$

So, as you can see, using basic differential rules, most (all?) of the "problems" with treating the derivative as a fraction go away, and, in its place, you have students who not only know what the notation means, they can use it effectively.

Disclaimer: I am currently working on a book on Calculus, the reason which I wrote it is largely to answer questions like these - most (all?) Calculus books have very awful discussions of differentials and what they mean.

• By the way, if anyone is interested in looking at a preview of my Calc book, see my profile page for contact information. – johnnyb Mar 5 '18 at 14:43
• Also, without providing justification, the proper way to write a partial derivative so that it can be used algebraically is $\frac{\partial_x u}{\mathrm{d}x}$, which is the partial derivative of $u$ with respect to $x$, with $x$ being the only variable allowed to freely/independently vary. – johnnyb Mar 5 '18 at 14:46
• If you don't mind sharing, what is your background? Over the years I've developed the impression that differentials (like infinitesimals) are historical artifacts from a less rigorous era of mathematics, and that their continued use today is due to engineers and physicists who like the intuition they provide, and don't mind the lack of rigor. Thus I rather disagree with the approach you are advocating. – Jonathan Mar 5 '18 at 15:18
• However, I'm curious why you think the infinitesimals lack rigor. There is nothing fundamentally non-rigorous about infinitesimals. I tend to use them intuitionally, but I do the same with counting numbers :) I'm not always sitting around thinking about sets and whatnot when I'm trying to add things. If you can think of a set of examples for which my approach doesn't work, but the standard approach does work, I'd appreciate it. If it is just extreme/obviously-edge edge cases, I don't think it would change my mind, but if you had a fairly standard case, I would seriously rethink my approach. – johnnyb Mar 5 '18 at 16:13
• In order to be rigorous, you need a precise definition. From this definition you need to prove the properties of differentials (such as the product rule). While this can be done, it seems more tedious than using the widely accepted definition of derivative. If your approach does not provide such definitions and proofs, then it is not considered mathematically rigorous. Which is not to say it does not work, or is not useful for students..... – Jonathan Mar 5 '18 at 17:58

My advice is to not worry too much about this but to teach the course and concepts and tricks and such instead. You have to have some nonrigorous understanding of something (in general) to understand it rigorously later. There is a reason we teach kids to count first versus Dedekind cuts. Richard Feynman wrote a nice essay on this issue of pedagogy.

Just mention the issue with the notation but don't super belabor it or distract from teaching the calculus itself. When you get to a problem where it matters to not have this confusion (as otherwise an error occurs), you can discuss it in the context of solving that problem or of correcting a false solution. But you are better off not spending super amounts of time "clarifying" this issue for people who don't even know how to take a derivative yet at all.

Calculus with analytic geometry by George Simmons addresses differentials and their treatment in the chapter 5 of the second edition.

• Greetings and welcome; unfortunately this answer does not have enough detail to stand on its own right now and will likely end up with a negative score or eventually deleted. To improve the answer, I recommend explaining what it is about the exposition in this book that makes you think it is superior to other books, with as many specifics as possible. – Chris Cunningham Dec 12 '19 at 18:24