Lebnit's notation question

  • Thread starter Juxt
  • Start date
8
0
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.:smile:
 

Tom Mattson

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Hi Juxt, and welcome to PF.

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

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
 
8
0
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

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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.
 
8
0
Thank you for that terrific analogy. What a lightbulb.
 
lol taking calculus in highschool rofl.
 

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