Why is the second derivative expressed differently in Lebnit's notation?

[Juxt]

I am in my first year of calculus at the high school level so be patient with me.

Why is it that in Lebnit's (spelling?) notation the second derivative is expressed as d²y/dx² ? My instructor did not know and from what we could work out it seems as if it should be expressed as d²y/d²x².

We have covered about what you have covered in the first eighth of a college level calculus course so try not to go too far over my head.

 

[quantumdude]

Hi Juxt, and welcome to PF.

The second derivative is just the first derivative of the first derivative.

Mathematically, the above sentence translates to:

d²y/dx²=(d/dx)(d/dx)y

You can loosely think of the two operators on the right as being multiplied like fractions to get:

(d²/dx²)y

or simply:

d²y/dx².

edit: fixed typo

 

[Juxt]

Maybe I was confused... is does dx= d times x or is dx like one variable (dx)? I am assuming that dx= d times x, as such I don't understand why it isn't d²y/d²x². Is my perception of the nomenclature wrong?

 

[quantumdude]

Originally posted by Juxt
Maybe I was confused... is does dx= d times x or is dx like one variable (dx)?

The second one.

dx is what you get when you take Δx(=x₂-x₁) and pass to the limit Δx-->0. Just as Δx is not Δ times x, so dx is not d times x.

 

[Juxt]

Thank you for that terrific analogy. What a lightbulb.

 

[PrudensOptimus]

lol taking calculus in high school rofl.