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Second Partial Derivatives of Implicit Functions

  1. Jul 13, 2005 #1
    I have been reviewing Calculus and have tripped up on figuring out to calculate the 2nd partial derivatives of imlicit functions. Kaplan and Spiegel give a cursory treatment to the subject in both of their "Advanced Calculus" books. Simply repeating the methods used to calculate the 1st derivatives doesn't appear to work. Any information would be appreciated.
     
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  3. Jul 13, 2005 #2

    Hurkyl

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    Why don't they appear to work?
     
  4. Jul 13, 2005 #3
    I don't know - perhaps because the equations are implicitly defined. I figured out how the answers were calculated, I can apply the same formula and get the correct answer, but I don't know why it is correct - the "naive" solution appears to work correctly, at least integrating the 2nd derivatives come about by using the "naive" solution produces the first. I am going to post in more detail later.
     
  5. Jul 14, 2005 #4
    Ok here it is. This example is from Kaplan's book - the detail is provided by me. Given a system of two functions:

    [tex]
    F \equiv ( x^2 + y^2 + u^2 + v^2 - 1)
    [/tex]
    [tex]
    G \equiv ( x^2 + 2y^2 - u^2 + v^2 - 1)
    [/tex]

    Where u and v depend on x and y:

    [tex]
    u = f(x,y)
    [/tex]
    [tex]
    v = g(x,y)
    [/tex]

    find

    [tex]
    \frac{\partial^2 x}{\partial u^2}
    [/tex]

    To find the first derivative, the method of Jacobians can be used. I am not going to provide the details, but the result is:

    [tex]
    \frac{\partial x}{\partial u} = - \frac{3u}{x}[/tex]

    Finding the second derivative depends on the fact that the differential operator can be chained:

    [tex]\frac{\partial}{\partial u} = \frac{\partial}{\partial x} \frac{\partial x}{\partial u}[/tex]

    So

    [tex]\frac{\partial^2 x}{\partial u^2} = \frac{\partial}{\partial u} \frac{\partial x}{\partial u} = \frac{\partial}{\partial u} (- \frac{3u}{x})[/tex]

    Considering the function as a product of two functions and applying the product rule for derivatives produces:

    [tex]\frac{\partial}{\partial u} [(- 3u)(\frac{1}{x})] = [\frac{\partial}{\partial u} (- 3u)](\frac{1}{x}) - 3u[\frac{\partial}{\partial u} (\frac{1}{x})] [/tex]

    After some manipulation:

    [tex]- \frac{3}{x} - 3u[\frac{\partial x}{\partial u}(\frac{\partial}{\partial x}\frac{1}{x})] = - \frac{3}{x} - 3u(- \frac{3u}{x})(- \frac{1}{x^2})[/tex]

    And finally, the answer:

    [tex]- \frac{3}{x} - \frac{9y^2}{x^3}[/tex]
     
    Last edited: Jul 14, 2005
  6. Jul 14, 2005 #5

    dextercioby

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    Are u using only the first equation (the one for F), or are u using the second (for G) as well?


    Daniel.
     
    Last edited: Jul 14, 2005
  7. Jul 14, 2005 #6
    Both F and G - they are used in the Jacobian to produce

    [tex]\frac{\partial x}{\partial u} = - \frac{3u}{x}[/tex].
     
  8. Jul 14, 2005 #7

    dextercioby

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    It's actually in reversed order

    [tex] \frac{\partial}{\partial u} = \frac{\partial x}{\partial u} \frac{\partial}{\partial x}[/tex]

    Daniel.
     
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