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Lebnit's notation question

  1. Nov 15, 2003 #1
    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:
     
  2. jcsd
  3. Nov 15, 2003 #2

    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
     
  4. Nov 15, 2003 #3
    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?
     
  5. Nov 15, 2003 #4

    Tom Mattson

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    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.
     
  6. Nov 15, 2003 #5
    Thank you for that terrific analogy. What a lightbulb.
     
  7. Nov 15, 2003 #6
    lol taking calculus in highschool rofl.
     
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