Lebnit's notation question

1. Nov 15, 2003

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.

2. Nov 15, 2003

Tom Mattson

Staff Emeritus
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

3. Nov 15, 2003

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?

4. Nov 15, 2003

Tom Mattson

Staff Emeritus
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.

5. Nov 15, 2003

Juxt

Thank you for that terrific analogy. What a lightbulb.

6. Nov 15, 2003

PrudensOptimus

lol taking calculus in highschool rofl.