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Differential and square of differential

  1. Nov 22, 2012 #1
    Hi
    I often see the following in books but I do not understand how they are equal. So can someone please tell me for what conditions does the following equality hold?
    ([itex]\frac{dy}{dx}[/itex]) 2 = (d2 y)/(dx2)
     
  2. jcsd
  3. Nov 22, 2012 #2
    No, generaly this equality doesn't hold.
    For example :
    y=x²
    dy/dx = 2x
    d²y/dx² = 2
    (dy/dx)² = (2x)² = 4x²
    2 is not equal to 4x²

    The equality (dy/dx)² = d²y/dx² holds only for one family of functions : y = - ln(ax+b).
    It doesn't hold for any other function.
     
  4. Nov 22, 2012 #3
    The equality that does hold is [itex]\displaystyle \left( \frac{d}{dx} \right)^2y = \frac{d^2y}{(dx)^2}[/itex], which is usually abbreviated (somewhat abuse of notation) as [itex]\displaystyle \frac{d^2y}{dx^2}[/itex].
     
  5. Nov 22, 2012 #4

    HallsofIvy

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    I can't help but wonder where you "often see" that? As others said, it is certainly NOT "generally" true. [itex]d^2y/dx^2= (dy/dx)^2[/itex] is a "differential equation". If we let u= dy/dx, we have the "first order differential equation" [itex]du/dx= u^2[/itex] which can be written as [itex]u^{-2}du= dx[/itex] and, integrating, [itex]-u^{-1}= x+ C[/itex] so that [itex]u= dy/dx= -\frac{1}{x+ C}[/itex]. Integrating that, [itex]y(x)= \frac{1}{(x+ C)^2}+ C_1[/itex]. Only such functions satisfy your equation.
     
  6. Nov 22, 2012 #5
    Integrating -1/(x+C) leads to -ln(x+C)+c = -ln(ax+b)
     
  7. Nov 22, 2012 #6
    He probably got confused and differentiated instead. Anyways, the identity given in the OP is not generally true, actually, it is almost surely false, given the family of functions that can be defined on R2.
     
  8. Nov 23, 2012 #7

    HallsofIvy

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    Yeah, I confuse very easily! Thanks.
     
  9. Nov 23, 2012 #8
    You do get sloppy very often, Halls. Shape up! :smile:
     
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