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Vague Partial Derivative

  1. Jun 2, 2012 #1
    Could someone please explain to me how to find the derivative of this:

    dy/dx = φ(x, y)

    Should I break up the equation to make it dy/dx = φ(x) + φ(y) and then derive the parts?

    I would then get d²y/dx² = ∂φ/∂x + ∂φ/∂y
    do I have to also multiply both terms by their respective derivatives of the inside variable?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 2, 2012 #2
    If φ(x, y) is arbitrary why do you think you can break it up to φ(x) + φ(y)? If φ(x, y) = xy, how can this be broken up into φ(x) + φ(y)?
     
  4. Jun 2, 2012 #3
    That's a good point. My professor wrote that the second derivative should be:

    ∂φ/∂x + ∂φ/∂y (dy/dx) = ∂φ/∂x + φ(∂φ/∂x)

    I've been trying to play around with the equation and see how I could get that answer.
    All of the partial derivatives I've done previously had equations that were equal to f(x,y) or such.
     
  5. Jun 2, 2012 #4
    Can you see how the Professor gets the left side? It's the chain rule.
     
  6. Jun 2, 2012 #5
    Yes, to take the second derivative of y, you should look at it as phi(x,y(x))

    so partial in x with respect to first entry, plus that with respect to second entry, which requires the chain rule.
     
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