What does the denominator in the second derivative mean?

Question

This occurred to me a few days ago.

We know that the derivative of a function $y=f(x)$ is $\frac{dy}{dx}$. This is because it represents how $y$ changes with $x$, which is the rate of change of $y$, or more specifically, the gradient of a function.

Then the second derivative is the rate of change of rate of change, or the rate of change of gradient. Since a general rate of change is $\frac{d}{dx}$, the second derivative is $(\frac{d}{dx})(\frac{dy}{dx})$. Thus, the expanded form is $\frac{d^2y}{dx^2}$.

My question is, is the denominator $d(x)^2$ or is it $(dx)^2$? Surely, it would be the latter, because when you expand $(\frac{d}{dx})(\frac{dy}{dx})$, the $(dx)(dx)$ would become $(dx)^2$. But then why is it never written with brackets? I'm sure that would confuse some people and I only realised it myself when I started thinking about the second derivative properly, in terms of what it actually means.

Answer 1

You are completely correct. When we write $\frac{d^2y}{dx^2}$, we really mean to write $\frac{d^2y}{(dx)^2}$, but those parentheses make the denominator difficult to read and write. Mathematicians accept the form $\frac{d^2y}{dx^2}$ without question, because it is not exactly an algebraic expression (although sometimes it can be treated as one), rather it is a notation that represents the concept of finding the rate of change of the rate of change.

Answer 2

You're right that it should really be $(dx)^2$ (if you were asking "why should that be?", see my answer here).

I suspect the brackets/parentheses aren't written because Leibniz himself didn't write them when he first presented the notation and everyone just followed his lead.

Answer 3

You are correct that $dx^2$ actually means $(dx)^2$. Or, even better, $(\text{d}x)^2$ (note the non-italicized $d$ because it is an operator not a variable). However, if you are treating differentials as algebraic units, you can't use the standard notation of $\frac{d^2y}{dx^2}$. If you actually perform the derivative of $\frac{dy}{dx}$ you will notice that $\frac{dy}{dx}$ is a quotient, and therefore requires the quotient rule for solving. After simplifying everything, the second derivative using algebraically-manipulable differentials is actually $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. Here, $d^2y$ is a shorthand for $\textrm{d}(\textrm{d}(y))$ and $dx^2$ is a shorthand for $(\textrm{d}x)^2$.

Using this notation, the differentials of the second derivative can be algebraically manipulated freely, while using the standard notation they cannot. The paper "Extending the Algebraic Manipulability of Differentials" explores this idea further.