# What does the denominator in the second derivative mean?

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.

• I find that thinking of it as notation rather than a rigorous representation is best, but this is a very interesting question, and I'm excited to see the responses. Jan 5, 2017 at 19:52
• Treating $dx$ as one symbol is quite common: the same happens when talking about infinitesimal quantities such as line elements. Jan 5, 2017 at 20:09
• Sidenote: under some interpretations, the chain rule says that $d(x^2)=2xdx$. (which is not what the second derivative is about)
– πr8
Jan 5, 2017 at 20:12
• Thank you everyone for your responses and for confirming my belief. @Mark S. I enjoyed reading your linked answer, it all made great sense and I now understand the whole concept a lot better. Thank you also pseudoeuclidean for the insight into how it is different from a typical algebraic expression, it now makes a lot more sense. Jan 5, 2017 at 21:02

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.

• When $dx$ is used to denote an infinitesimal either in Leibniz's slightly informal work or in nonstandard analysis (with an implied standard-part function) it pretty much is an algebraic expression. Jan 5, 2017 at 21:37

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.

• By the way, can't $(\mathrm dx)^2$ also be interpreted as the $2$-tensor field $\mathrm dx\otimes\mathrm dx$ ? Apr 9, 2020 at 12:47
• @Maximilian I'd have to review, but my gut says "in an integral, sure. But not in the fraction under discussion here." Apr 9, 2020 at 16:55

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.