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Chain Rule Question

  1. Oct 25, 2010 #1
    1. The problem statement, all variables and given/known data Reduce the order of a Cauchy-Euler Equation



    2. Relevant equations [itex] x = e^t \mbox{ and } \ln x = t [/itex]



    3. The attempt at a solution
    [itex] \displaystyle \frac{d y}{d x} = \displaystyle \frac{d y}{d t} \displaystyle \frac{d t}{d x} = \displaystyle \frac{d y}{d t} \cdot \displaystyle \frac{1}{x}
    [/itex]
    and thus
    [itex]
    \displaystyle \frac{d^2 y}{d x^2} = \displaystyle \frac{d y}{d t} \cdot \displaystyle \frac{-1}{x^2} + \displaystyle \frac{1}{x} \displaystyle \frac{d}{d x} \Bigl ( \displaystyle \frac{d y}{d t} \Bigl )
    [/itex]


    Here is where I am getting stuck, specifically on [itex] \displaystyle \frac{d}{d x} \Bigl ( \displaystyle \frac{d y}{d t} \Bigl )
    [/itex] this step. I know what I should get...
    [itex]
    \displaystyle \frac{1}{x} \Bigl ( \displaystyle \frac{d^2 y}{d t^2} \cdot \displaystyle \frac{1}{x} \Bigl )
    [/itex]
    But uhhh not getting it. Thanks in advance!
     
  2. jcsd
  3. Oct 25, 2010 #2

    vela

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    You apply the chain rule again, except instead of applying it to y, you're applying it to dy/dt. It's a bit clearer if you look at it in terms of operators:

    [tex]\frac{d}{dx} f = \frac{dt}{dx}\frac{d}{dt} f = \frac{1}{x}\frac{d}{dt} f[/tex]

    so

    [tex]\frac{d}{dx} = \frac{1}{x} \frac{d}{dt}[/tex]
     
  4. Oct 25, 2010 #3
    Ahhh...very nice! Thanks a bunch!
     
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