# Lebnit's notation question

#### 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.

#### Tom Mattson

Staff Emeritus
Gold Member
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?

#### Tom Mattson

Staff Emeritus
Gold Member
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 &Delta;x(=x2-x1) and pass to the limit &Delta;x-->0. Just as &Delta;x is not &Delta; times x, so dx is not d times x.

#### Juxt

Thank you for that terrific analogy. What a lightbulb.

#### PrudensOptimus

lol taking calculus in highschool rofl.

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