usage of Leibniz notation for things like $\frac{d^2y}{dt^2}$ and $\frac{dy'}{dy}$

Question

I've read the other posts on this site about whether you can treat $\frac{dy}{dt}$ as a fraction. There are a lot of conflicting opinions, but many seem to be saying that treating it as a fraction works fine for single-variable calculus as long as you don't write anything obviously counterproductive, like $$\left( \frac{dy}{dx} \right)^2 = \frac{(dy)^2}{(dx)^2} \ \ \text{ or } \ \ 2^{dy/dx} = \sqrt[dx]{2^{dy}}.$$

I've also read that $y''(t)$ can be written as $\frac{d^2 y}{dt^2}$ with the reasoning that $$y''(t) = \frac{d\frac{dy}{dt}}{dt} = \frac{d}{dt}\frac{dy}{dt} = \frac{d^2 y}{(dt)^2} = \frac{d^2 y}{dt^2}$$

This seems to make sense until I consider an expression like $\frac{dy'}{dy}$ (the derivative of $y'(t)$ with respect to $y$). If I work with Leibniz notation in the same way as before, I get $$\frac{dy'}{dy}=\frac{d\frac{dy}{dt}}{dy}=\frac{d}{dy}\frac{dy}{dt}=\frac{d}{dt}\frac{dy}{dy}=\frac{d}{dt}1=0$$

But from my understanding of derivatives and the chain rule, it makes sense to say $$\frac{d\frac{dy}{dt}}{dy} \equiv \frac{d^2y}{dt^2}\frac{dt}{dy}$$ which is not necessarily zero.

My question is, are there any consistent rules for how to use Leibniz notation? Maybe you have to treat $\frac{d}{dt}$ and $\frac{dy}{dt}$ as "atomic" even in single-variable calculus, which would mean the $\frac{d^2y}{dt^2}$ notation is extremely misleading. Or am I doing/understanding something wrong here?

Edit: Here's one last example of confusing Liebniz notation. A seemingly reasonable but incorrect justification for the 2nd derivative chain rule: $$\frac{d^2y}{(dx)^2} = \frac{d^2y}{(dt)^2}\frac{(dt)^2}{(dx)^2} = \frac{d^2y}{(dt)^2}(\frac{dt}{dx})^2$$ The correct but very notationally confusing 2nd derivative chain rule: $$\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}(\frac{dx}{dt})^2+\frac{dy}{dx}\frac{d^2x}{dt^2}$$

Answer 1

The justification you gave for denoting the expression $y''(dt)^2$ by $d^2y$ is not really a proof but rather a heuristic argument, and therefore unreliable in general (as you yourself have illustrated). Viewing the derivative and the second derivative as quotients is possible in the context of a framework that allows for infinitesimals, such as Robinson's infinitesimal analysis. Before we get to the second derivative, let us see how using $\frac{dy}{dt}$ for the first derivative is justified.

We start with an infinitesimal increment $\Delta t$ of the independent variable $t$, and form the corresponding change of the dependent variable $y$ by setting $\Delta y=f(t+\Delta t)-f(t)$. Then we can form the ratio of infinitesimals $\frac{\Delta y}{\Delta t}$ but in general it will not be the derivative but only infinitely close to it. Thus we define the derivative as the standard part (or shadow) of the ratio $\frac{\Delta y}{\Delta t}$ (when it exists). We then define a new dependent variable $dy$ by setting $dy=f'(t)dt$, where $dt=\Delta t$ (for the independent variable, there is no difference between $\Delta t$ and $dt$). Then $\Delta y$ and $dy$ differ by an infinitesimal term that's negligible compared to $\Delta y$.

For second derivative, the pertinent observation is that one can define $f''(t)$ (when it exists) as the standard part of the ratio $\frac{f(t+\Delta t) +f(t-\Delta t) -2f(t)}{\Delta t^2}$. In this sense it is a "second difference" (to use a Leibnizian term). This motivates the definition of $d^2y$ as the expression $f''(t)(\Delta t)^2$.

Incidentally, writing $\left( \frac{dy}{dx} \right)^2 = \frac{(dy)^2}{(dx)^2}$ or $2^{dy/dx} = \sqrt[dx]{2^{dy}}$ is perfectly legitimate in the context of Robinson's infinitesimal analysis, by the transfer principle.

Note that the formula for the chain rule, $\frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}$ is not proved by formal cancellation of $dx$ in the numerator and the denominator; rather, careful proof is required. The problem here is that the $dx$ in the denominator is an independent variable whereas in the numerator is a dependent variable, so they can't be canceled without further argument. Similarly, in the formula for the second derivative obtained by applying the Leibniz rule to the chain rule formula, in the expression $\frac{d^2y}{dx^2}\big(\frac{dx}{dt}\big)^2$ the $dx$ in the numerator and the $dx$ in the denominator cannot be canceled, because $dx$ in the denominator is an independent variable and the $dx$ in the numerator is a dependent variable (i.e., dependent on $t$). In any piece of mathematical notation, one has to strike a balance between being concise, on the one hand, and being explicit, on the other. Normally one would have to write $dy(x,dx)$ in place of $dy$ and such to be completely explicit, but then the formulas will be harder to read.

At any rate, paradoxes arise only if one insists of trying to apply formal manipulations of symbols without thinking of their meaning.

Answer 2

For higher-order differentials to work algebraically, you need to adopt a notation that is a little non-standard. The typical notation, $\frac{d^2y}{dx^2}$ does not allow for algebraic manipulations.

However, if you take the first derivative, $\frac{dy}{dx}$ and take the derivative of it, you will notice that it is indeed a quotient, and therefore needs to use the quotient rule. The result will be that the notation for the second derivative is $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. This is fully algebraically manipulable. For instance, by straightforward algebraic notations, you can establish an inverse function theorem for the second derivative. Because the second derivative of $x$ with respect to $y$ is (using this notation) $\frac{d^2x}{dy^2} - \frac{dx}{dy}\frac{d^2y}{dy^2}$, we can see how to get there as follows:

$$ y'' = \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2} \\ y''\, dx^3 = d^2y\,dx - dy\,d^2x \\ y''\, \frac{dx^3}{dy^3} = \frac{d^2y}{dy^2}\frac{dx}{dy} - \frac{d^2x}{dy^2} \\ -y''\,\frac{dx^3}{dy^3} = \frac{d^2x}{dy^2} - \frac{d^2y}{dy^2}\frac{dx}{dy} \\ -y'' \frac{1}{y'}^3 = x'' $$

This is more fully fleshed out in two papers I was involved in, "Extending the Algebraic Manipulability of Differentials" and "Total and Partial Differentials as Algebraically Manipulable Entities". But, in short, there is no problem in manipulating differentials algebraically as long as you use the proper notation, with notation being especially important for higher-order differentials and partial differentials. Most of the theoretical objections have come from general objections to infinitesimals, but the hyperreal number system has demonstrated that there is no logical problem with such numbers.

To address your specific example, though, the problem is that you are treating $d$ by itself as an entity. It is not. It is an operator. $dx$ is actually $d(x)$. You can think about it kind of like a function, such as $\sin(x)$, at least when considering parenthesis movement. The problem in your example is that you haven't properly parenthesized it. I will expand parentheses here so it is easy to spot the problem:

$$\frac{dy'}{dy} = \frac{d(y')}{d(y)} = \frac{d\left(\frac{d(y)}{d(x)}\right)}{d(y)}$$

As you can see, you can't get the $dy$s to cancel out because one is inside the other differential. Now, we can compute the differential, which yields:

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

So, as you can see, you can do manipulation, it's just that you have to keep in mind the type of operations you are actually doing and what they are being applied to. Similar to functions, you can't just willy-nilly move the argument of an operator inside or outside the operator without justification.

Edit: OP asked for elaboration on the chain rule for the second derivate

For the chain rule for the second derivative, the chain rule still "works" but it is not necessary. In other words, the chain rule simply tells you which algebraic manipulations will bring you from one derivative to another.

So, let's look at what the chain rule is (we will use Arbogast/Euler D-notation to avoid confusion):

$$ D^2_t(y) = D^2_x(y)\,(D^1_t(x))^2 + D^1_x(y)\,D^2_t(x) $$

So, the left side is $\frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2t}{dt^2}$. So, let's just do the algebra of the right-hand side and see what happens. I'll over-parenthesize so you can more easily see the components.

$$ \left(\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}\right)\, \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dx}\right)\,\left(\frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2}\right) \\ = \left(\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}\right)\, \left(\frac{dx^2}{dt^2}\right) + \left(\frac{dy}{dx}\right)\,\left(\frac{d^2x}{dt^2} - \frac{dx}{dt}\frac{d^2t}{dt^2}\right) \\ = \left(\frac{d^2y}{dt^2} - \frac{dy}{dx}\frac{d^2x}{dt^2}\right) + \left(\frac{dy}{dx}\frac{d^2x}{dt^2} - \frac{dy}{dt}\frac{d^2t}{dt^2}\right) \\ = \frac{d^2y}{dt^2} + \left(-\frac{dy}{dx}\frac{d^2x}{dt^2} + \frac{dy}{dx}\frac{d^2x}{dt^2}\right) - \frac{dy}{dt}\frac{d^2t}{dt^2} \\ = \frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2t}{dt^2} $$

So, you can see, the chain rule for the second derivative simply encodes the steps we would need to algebraically manipulate one derivative into the other.

Answer 3

Leibniz notation is consistent, and even more I would say, it is useful by allowing shortcuts and permitting to carry computations with lightened notations. Nonetheless, it necessitates some practice and you will get used to it.

Unfortunately, there is no real set of rules ensuring consistency, or perhaps there exists a single one implicitly : each step has to be justified. In the concrete case you presented, the "wrong move" lies in the equality $D_yD_ty = D_tD_yy$, since the (total) derivatives with respect to $y$ and $t$ don't commute, because $y$ depends on $t$.

Also, I wouldn't pay too much attention to the paper you linked (see comment below). Indeed, its authors make non-standard manipulations. If I have understood them well, they try to redefine the second-order derivative by considering the first-order derivative $y'$ as a bivariate function of $x$ and $y$, i.e. $y' = y'(x,y)$, before applying the chain rule (because $y$ itself depends on $x$). However, they end up defining the second-order derivative $y''$ from itself and a useless vanishing term (because $x'' = 0$) in order to tally with a chain rule structure. In consequence, they haven't proven anything, apart from rewriting $y''$ in Leibniz style.

Finally, it is to be noted that higher-order derivatives in Leibniz notation are commonly treated as differentials, such as in the case of the second-order variation of a bivariate function $f(x,y)$ for example : $$ \mathrm{d}^2f(x,y) = \frac{\partial^2f}{\partial x^2}\mathrm{d}x^2 + \frac{\partial^2f}{\partial x\partial y}\mathrm{d}x\mathrm{d}y + \frac{\partial^2f}{\partial y^2}\mathrm{d}y^2 $$ Then you can divide this expression by a "second-order infinitesimal", i.e. $\mathrm{d}x^2$, $\mathrm{d}y^2$ or $\mathrm{d}x\mathrm{d}y$ in order to select the suitable term. You may even do it with respect to other parameters, such as $\mathrm{d}t^2$ or $\mathrm{d}s\mathrm{d}t$, which will induce the chain rule automatically if $x,y$ depend on them.