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Change of variables/total differential derivation

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data


    http://gyazo.com/e7dc83f3f93d01391663d02bf92c0b26

    2. Relevant equations

    I am trying to derive equation 6 from 5 using 1, but I haven't got anywhere. This isn't a homework question, just an information sheet. Can someone show me the steps (no jumps please) on obtaining equation 6. After that, for practice I will attempt df/dv

    Thanks
    :)
     
  2. jcsd
  3. Apr 2, 2012 #2

    SammyS

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    It's pretty basic.

    Group the terms in (5):
    First distribute [itex]\displaystyle \left.\frac{\partial f}{\partial x}\right|_y\text{ and }\left.\frac{\partial f}{\partial y}\right|_x\ .[/itex]θ

    Then factor out du & dv .

    [itex]\displaystyle df=\left(
    \left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial u}\right|_v
    +\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial u}\right|_v
    \right)du +\left(
    \left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial v}\right|_u
    +\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial v}\right|_u
    \right)dv
    [/itex]



     

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  4. Apr 3, 2012 #3
    I can easily get to the second equation
    [itex]\displaystyle df=\left(
    \left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial u}\right|_v
    +\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial u}\right|_v
    \right)du +\left(
    \left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial v}\right|_u
    +\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial v}\right|_u
    \right)dv
    [/itex]

    but how does that become equation 6? We've gone from 4 terms to 6 and remove dv and du

    Thanks
     
  5. Apr 3, 2012 #4

    SammyS

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    All that equation 6 gives is the coefficient of du.

    The coefficient of dv is similar.

    Compare equation 1 with [itex]\displaystyle df=\left(
    \left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial u}\right|_v
    +\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial u}\right|_v
    \right)du +\left(
    \left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial v}\right|_u
    +\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial v}\right|_u
    \right)dv\ .[/itex]
     
  6. Apr 4, 2012 #5
    Awesome. I see :)
     
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