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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.

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    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. – 2017-01-05
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    Treating $dx$ as one symbol is quite common: the same happens when talking about infinitesimal quantities such as [line elements](https://en.wikipedia.org/wiki/Line_element). – 2017-01-05
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    Sidenote: under some interpretations, the chain rule says that $d(x^2)=2xdx$. (which is not what the second derivative is about) – 2017-01-05
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    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. – 2017-01-05

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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.

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    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. – 2017-01-05
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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.