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Homework Help: Partial derivative of radial basis function

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the partial derivatives (∂f/∂x & ∂f/∂y)

    2. Relevant equations


    3. The attempt at a solution

    really confusing me with the use of the summation and power to 3/2. This is my attempt, most definitely wrong but still tried.

    ∂f/∂x = x + c1*(2*(x-x1))*([( x-x1 )^2 + (y-y1)^2)]^(1/2)) + ... + cN*(2*(x-xN))*([( x - xN )^2 + (y-yN)^2)]^(1/2))

    ∂f/∂y = y + c1*(2*(y-y1))*([( x-x1 )^2 + (y-y1)^2)]^(1/2)) + ... + cN*(2*(y-yN))*([( x - xN )^2 + (y-yN)^2)]^(1/2))

    sorry, I probably should know this, but I can't really think straight right now -_-
    not sure if its just double chain rule or what...

    thank you for any help
  2. jcsd
  3. Nov 3, 2011 #2


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    Homework Helper

    looks close
    f(x,y) = a + b_1x + b_2y + \sum_{j=1}^{N}c_j((x-x_j)^2+(y-y_j)^2)^{\frac{3}{2}}

    there's no problem taking the partial derivative under the sum, so as you say it just becomes a chain rule question
    f_x = \frac{\partial}{\partial x}f(x,y) = \frac{\partial}{\partial x}(a + b_1x + b_2y + \sum_{j=1}^{N}c_j((x-x_j)^2+(y-y_j)^2)^\frac{3}{2}

    f_x = b_1 + \sum_{j=1}^{N}\frac{\partial}{\partial x}c_j((x-x_j)^2+(y-y_j)^2)^\frac{3}{2}

    using the chain rule once gives
    f_x = b_1 + \sum_{j=1}^{N}\frac{3}{2}c_j((x-x_j)^2+(y-y_j)^2)^\frac{1}{2} \frac{\partial}{\partial x}((x-x_j)^2+(y-y_j)^2)

    second time
    f_x = b_1 + \sum_{j=1}^{N}\frac{3}{2}c_j((x-x_j)^2+(y-y_j)^2)^\frac{1}{2} 2(x-x_j)\frac{\partial}{\partial x}(x-x_j)
  4. Nov 3, 2011 #3
    Much appreciated lanedance, cheers
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