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So, if we think of the higher order derivatives as the $\frac{d}{dx}$ operator to the power of the order multiplied by y, $\left[\frac{d^n y}{d x^n} = \left(\frac{d}{dx}\right)^n(y)\right]$

Trying this on the second derivative,

$$=\left(\frac{d}{dx}\right)^2(y)$$ $$=\left(\frac{d}{dx}\right)\left(\frac{d}{dx}\right)(y)$$ $$=\left(\frac{d·d}{dx·dx}\right)(y)$$ $$\neq \left(\frac{d^2}{d^2x^2}\right)(y)$$ but rather $$=\left(\dfrac{d^2}{dx^2}\right)(y)$$
Why is the product of the denominator, $(dx)(dx)$, not $d^2x^2$ or $(dx)^2$?
Does $dx^2$ represent $d^2x^2$ because it's weird to square the $d$? If so, why is the $d$ in the numerator squared?

A similar question to mine was asked here, but none of the answers answered my question about the denominator, or at least I did not understand them if they did.

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Something important about Leibniz notation that is often ignored is that $\frac{d}{dx}$ is something you apply to a function to differentiate it. What you are not doing is multiplying an object d to a function f and dividing by d applied to some other function/variable x. The dx in the denominator should be thought of as its own object. Because of this a secondary application of $\frac{d}{dx}$ results in $\frac{dd}{dxdx}\neq\frac{dd}{ddxx}$ which because it is cumbersome has been abbreviated as $\frac{d^2}{dx^2}$ however because the dx is one object $dx^2\equiv (dx^2)$.

This does showcase the ambiguity of Leibniz notation however since $\frac{d^2}{dx^2}$ is a second derivative with respect to the variable x while $\frac{d}{dx^2}$ is a first derivative with respect to the function $x^2$.

Note that many people who have studied differential forms think that the "differentials" dx and df are actually differential one forms, however this is not the case since a differential one form squared is identically 0. $dx\wedge dx=d^2x=0$

Sam
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    Did you mean to write $dx^2\equiv(dx)^2$ instead of $dx^2\equiv(dx^2)$? – bof Jul 03 '19 at 18:28
  • @bof No I meant exactly what I wrote because I am saying dx is a single mathematical object. However your comment brings up a good point about the ambiguity of Leibniz notation since $dx^2$ could be interpreted as differentiating with respect to $x^2$ instead of a second derivative. – Sam Jul 03 '19 at 18:30