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Differential operators - the rules

  1. Apr 21, 2010 #1
    I always get slightly confused with the rules of differentials.

    now [tex]\frac{d^{2}y}{dx^{2}}[/tex] is the scond derivative of the function y(x

    but rooting this does NOT give the first derivative dy/dx

    However, with the operator [tex]\frac{d^{2}}{dx^{2}}[/tex], it seems that you can root this and it DOES give the first derivative.

    Can someone please explain this to me? I may be wrong, but this seems to be the case in my quantum mechanic notes
  2. jcsd
  3. Apr 22, 2010 #2


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    Hi randybryan! :wink:

    Yes, d2y/dx2 = (d/dx)(d/dx)y, but ≠ [(d/dx)y]2.

    It's for the same reason that sin(sin(y)) ≠ [sin(y)]2 :smile:
  4. Apr 22, 2010 #3


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    I suspect that your "quantum mechanic notes" (don't hold mathematicians responsible for what a physicist says!:wink:) are using a special notation in which "square root" has some kind of operational definition. That is, "[itex]f^2(x)[/itex]" does not mean f(x) times itself but f(f(x)) and [itex]\sqrt{f}[/itex] is the inverse of that.
    Last edited by a moderator: Apr 23, 2010
  5. Apr 22, 2010 #4
    Ahh the ongoing feud that is Maths nomenclature vs Physics nomenclature.
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