# 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

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.

• 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