Question

Why isn't the notation for the second derivative `dy^2/dx`?

Answer 1

See below.

Explanation:

Probably because `(dy^2)/dx` would be read as the derivative of `y^2` in respect of `x`.

The general notation `(d^2y)/dx^2` could be misconstrued as the derivative in respect of `x^2`, but then you can find lots of flaws in mathematical notations, but we just have to accept that this is how they have been defined.

Answer 2

Please see below.

Explanation:

For `f(x)` a function of `x`, we can denote the derivative in several ways.

On family of notations uses a "differential operator".
This is a notation that tells us "take the derivative".

For example `D_x(5x^2+3)` means "The derivative, with respect to `x`, of the function `f(x)=5x^2+3`. SO `D_x(5x^2+3) = 10x`.

`d/dx` is another notation for "the derivative with respect to `x` of . . . #

So `d/dx(5x^2+3)` means the same thing as `D_x(5x^2+3)`.

To take the second derivative, we take the derivative twice:

`D_x(D_x(5x^2+3))` is written `D_x^2(5x^2+3)`

`d/dx(d/dx(5x^2+3))` could, perhaps, be written `(d/dx)^2(5x^2+3)`, but the standard notation is `d^2/dx^2(5x^2+3)`.

If we let `y = 5x^2+3`, then these examples look like:

First derivative: `D_x(y)` and `d/dx(y)` which is also written `dy/dx`.

Second derivative: `D_x^2(y)` and `d^2/dx^2(y)` which is also written `(d^2y)/dx^2`.

In this notation, we do not think of `dx` as `d` times `x`

Instead, we are thinking of `dx` as a single quantity. That is why we do NOT write `d^2/(dx)^2(y)`

Additional Note

I would interpret `dy^2/dx` as `d/dx(y^2)`. Using the chain rule, I would evaluate to get `d/dx(y^2) = 2y * dy/dx`.

Using `y=5x^2+3` again, we have `d/dx(5x^2+3)^2 = 2(5x^2+3)*5 = 10(5x^2+3)`

Note 2

`(dy/dx)^2` means the square of the derivative of `y`.
Using `y=5x^2+3` again, we have

`(dy/dx)^2 = (10x)^2 = 100x^2`.

This kind of expression can come up if using implicit differentiation to find 2nd or higher derivatives. (If we don't pause to solve for `dy/dx` before our second differentiation.)

Answer 3

See below.

Explanation:

`d/(dx)(dy/dx) = (d^2y)/(dx^2)`
`d/(dx)(d/dx)y = (d^2)/(dx^2)y = (d^2y)/(dx^2)`