Explaining intuitively the notation for derivatives

Question

I am not clear on the notation we use for the derivative.
So let's assume that $y=f(x)=6x-x^2$ so the first derivative is $$\frac{Δy}{\Delta x}=\frac{dy}{dx}=6-2x=F(x)$$
The notation for the second derivative is:
$$\frac{dF(x)}{dx}=\frac{d^2y}{dx^2}=-2$$
My question is about the notation of the second derivative. Does the superscript 2 indicate to the power of 2 or just an order?
Because I can see that:
$$\frac{dF(x)}{dx}=\frac{d}{dx}\frac{dy}{dx}$$
But is that a multiplication really? Because dy is just y2-y1 and what is the plain d then?

Update based on comments
Why is it not $\frac{d^2y}{d^2x^2}$ if the lower part is multiplication?
I can see that the upper part is the differentiation applied twice since it is dF(x)=ddy and seems to indicate just a second (order) differentiation but I am not sure what the denominator indicates. Why is it $\frac{d^2y}{dx^2}$ and not $\frac{d^2y}{d^2x^2}$? What does the $x^2$ really mean?

Answer 1

Short answer

There are good reasons for a notation like $\dfrac{d^2y}{(dx)^2}$, but unfortunately the convention happens to be to drop the parentheses in the denominator. This goes back to Leibniz's original notation.

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Long answer

(Note: I use $\mathrm{d}$ instead of the italic $d$ so I never forget that $\mathrm{d}$ is not like a variable.)

$\Delta$ and $\mathrm{d}$ are not quite the same thing, and understanding them well may help you understand what's going on with the second derivative, so I'll start there (but you may be able to skip the first section or two).

I. The meanings of $\Delta$ and $\Delta y/\Delta x$

$\Delta$ generally means something like "a difference/change in". $\Delta x$ would often be a number that represents how much $x$ changed by (say, between two points of interest). If $y$ is a quantity that depends on $x$, then $\Delta y$ means something like "a change in $y$ (and since $y$ depends on $x$ we know this change will depend on how much we're changing $x$ by: i.e. it'll change based on $\Delta x$)". For example, if $y$ stands for $f(x)$, then it would be reasonable to write $\Delta y=f(x+\Delta x)-f(x)$.

The slope of a secant line (sometimes called "the average rate of change of $y$ with respect to $x$") between the points $\left(x,f(x)\right)$ and $\left(x+\Delta x,f\left(x+\Delta x\right)\right)$ is then given by $$\dfrac{f\left(x+\Delta x\right)-f(x)}{\left(x+\Delta x\right)-x}=\dfrac{f\left(x+\Delta x\right)-f(x)}{\Delta x}=\dfrac{\Delta y}{\Delta x}\tag{1}$$

II. The meanings of $\mathrm d$ and $\mathrm dy/\mathrm dx$

$\mathrm d$ generally means something like "an infinitesimal/tiny difference/change in". If $y$ is a quantity that depends on $x$, then $\mathrm dy$ means something like "a tiny change in $y$ (and since $y$ depends on $x$ we know this change will depend on how much we're changing $x$ by: i.e. it'll change based on $\mathrm dx$)".

Unfortunately $\mathrm dy$ isn't very useful on its own since it's infinitesimal, but we can still compare tiny changes in $y$ to tiny changes in $x$ to get a sense for how $y$ grows.

The slope of a tangent line (sometimes called "the instantaneous rate of change of $y$ with respect to $x$") at the point $\left(x,f(x)\right)$ should be what slopes of secant lines approach as $\Delta x$ gets small, so we use a limit. We define $\dfrac{\mathrm dy}{\mathrm dx}={\displaystyle \lim_{\Delta x\to0}}\dfrac{\Delta y}{\Delta x}$. This way, we have

$$\dfrac{\mathrm dy}{\mathrm dx}=\lim_{\Delta x\to0}\dfrac{f\left(x+\Delta x\right)-f(x)}{\Delta x}=\lim_{h\to0}\dfrac{f\left(x+h\right)-f(x)}{h} \tag{2}$$

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III. The meaning of $\Delta^2 y/\left(\Delta x\right)^2$

Sometimes we're interested in changes of changes (like "acceleration" as the change of the change of position with respect to time). As usual, we'll assume that $y$ stands for $f(x)$. To make things easier to write/think about, I'll give the function name $g(x)$ to $\Delta y$, so that $g(x)=f\left(x+\Delta x\right)-f(x)$. Note that $g(x)$ technically depends on $\Delta x$, but I don't want to write something like $g_{\Delta x}$ every time.

Now we can calculate some things and see where they lead: First, we can calculate the change in the change in $y$:

\begin{align}\Delta\left(\Delta y\right)&=\Delta\left(g\left(x\right)\right) \\&=g\left(x+\Delta x\right)-g\left(x\right) \\&=\left(f\left(\left(x+\Delta x\right)+\Delta x\right)-f\left(x+\Delta x\right)\right)-\left(f\left(x+\Delta x\right)-f(x)\right) \\&=f\left(x+2\Delta x\right)-2f\left(x+\Delta x\right)+f(x)\tag{3}\end{align}

What about the difference in the slopes of two secant lines?: \begin{align}\Delta\left(\dfrac{\Delta y}{\Delta x}\right) & =\Delta\left(\dfrac{g(x)}{\Delta x}\right)\\ & =\dfrac{g\left(x+\Delta x\right)}{\Delta x}-\dfrac{g(x)}{\Delta x}\\ & =\dfrac{g\left(x+\Delta x\right)-g(x)}{\Delta x}\\ & =\dfrac{\Delta\left(g(x)\right)}{\Delta x}\\ & =\dfrac{\Delta\left(\Delta y\right)}{\Delta x}\tag{4} \end{align}

Finally, we can look at how the slopes of secant lines are changing as compared to the change in $x$: $$\dfrac{\Delta\left(\dfrac{\Delta y}{\Delta x}\right)}{\Delta x}=\dfrac{\dfrac{\Delta\left(\Delta y\right)}{\Delta x}}{\Delta x}=\dfrac{\Delta\left(\Delta y\right)}{\left(\Delta x\right)^{2}}\tag{5}$$

Because it resembles multiplication, it's common to write $\Delta\left(\Delta y\right)$ as $\Delta^2 y$, so that this quantity we just looked at is $\dfrac{\Delta^2 y}{\left(\Delta x\right)^{2}}$.

IV. The meaning of $\mathrm d^2 y/\mathrm dx^2$

Once we have a derivative, like $\dfrac{\mathrm dy}{\mathrm dx}$, we might want to differentiate it again to find things like "instantaneous acceleration". So we're interested in the following: \begin{align}\dfrac{\mathrm{d}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}{\mathrm{d}x} & =\lim_{\Delta x\to0}\dfrac{\Delta\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}{\Delta x}\\ & =\lim_{\Delta x\to0}\dfrac{\Delta\left({\displaystyle \lim_{\Delta x\to0}\dfrac{\Delta y}{\Delta x}}\right)}{\Delta x}\\ & =\lim_{\Delta x\to0}\dfrac{\Delta\left({\displaystyle \dfrac{\Delta y}{\Delta x}}\right)}{\Delta x}\tag{$\star$}\\ & =\lim_{\Delta x\to0}\dfrac{\Delta^{2}y}{\left(\Delta x\right)^{2}}\tag{6} \end{align}

Answer to the OP

By analogy with $\dfrac{\mathrm{d}y}{\mathrm{d}x}$, it's reasonable to write the above quantity as $\dfrac{\mathrm{d}^{2}y}{\left(\mathrm{d}x\right)^{2}}$. To save writing of parentheses, the convention is often to write $\dfrac{d^{2}y}{dx^{2}}$, which I think can be confusing.

Technical note

When we went to the line denoted with ($\star$) above, I cheated a bit. When we're talking about limits, maybe the order or relative speeds of the two limits could matter. It would probably be better to have two different $\Delta x$s with their own limits. But for nice enough functions (at least functions with a continuous second derivative), this doesn't matter and it's safe to treat the inner and outer $\Delta x$s as the same.

Answer 2

Using differentials in this sense is really the act of a function. I learned from a video (I believe by Edward Burger) that you should think of $\frac{d}{dx}$ as a verb, to differentiate the thing, and $\frac{dy}{dx}$ as a noun, the derivative of $y$.

We may decompose higher order derivatives by peeling, like sums. For example:

$$ \frac{d^2 y}{dx^2} = \frac{d}{dx} \frac{dy}{dx} $$.

The operator $\frac{d}{dx}$ tells you to differentiate $\frac{dy}{dx}$, the derivative of $y$. It's like function notation, how we tell students that $f(x)$ does not mean $f$ times $x$.