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Variable transformation in a derivative

  1. Jun 27, 2012 #1
    Hi

    Maybe I don't see the wood because of all the trees, but:

    You have a second derivative [itex] \frac{\mathrm{d}^2}{\mathrm{d}x^2} e^{-ax} \cdot u(ax) [/itex]

    Now you make the variable transformation [itex] w=ax [/itex]

    How to express

    [itex] \frac{\mathrm{d}^2}{\mathrm{d}w^2} [/itex]

    Thanks
    Greetings
     
  2. jcsd
  3. Jun 27, 2012 #2

    mathman

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    Your question is confusing. Are you substituting w = ax in the original function or after you have the second derivative in x?
     
  4. Jun 27, 2012 #3

    HallsofIvy

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    If you let w= ax, then [itex]f(x)= e^{-ax}u(ax)[/itex] becomes [itex]f(w)= e^ww[/tex]. Of course, then, [itex]df/dw= e^ww+ e^w= e^w(w+1)[/itex] and the [itex]d^2f/dw^2= e^w(w+ 1)+ e^w= e^w(w+ 2)[/itex].

    Nothing unusual about that. Note, however, that while [itex]df/dx= (df/dw)(dw/dx)[/itex], it is NOT true that "[itex]d^2f/dx^2= (d^2f/dw^2)(dw/dx)[/itex]. Rather, [itex]d^2f/dx^2= d/dx(df/dx)= d/dx((df/dw)(dw/dx))= d/dx(df/dw)+ (df/dw)(d^2w/dx^2)[/itex][itex]= (d/dw(df/dw))(dw/dx)+ (df/dw)(d^2w/dx^2)=(d^2f/dw^2)(dw/dx)+ (df/dw)(d^2w/dx^2)[/itex].
     
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