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Homework Help: Implicit Function

  1. May 20, 2008 #1
    I thought I had this but in doing the problem I realized I didn't (or maybe I didn't).
    One of the problems was that in class and notes all examples were done in terms of f(x,y). Obviously in h/w, the problem is given as f(x1,x2,x3) - just to confuse me !!

    Problem statement

    a- For the following equation, decide whether they implicitly determine functions near 0
    b- If the equation implicitly determine a function, compute the partial derivatives at 0


    F1(x) = ( x1[tex]^{2}[/tex] + x2[tex]^{2}[/tex] + x3[tex]^{2}[/tex] )[tex]^{3}[/tex] - x1 + x3 = 0;
    F2(x) = cos (x1[tex]^{2}[/tex] + x2[tex]^{4}[/tex]) + exp(x3) - 2 = 0


    For Part a-

    I know I have to compute the determinant of a matrix. So for this example I played out with x1,x2,x3 and saw which determinant would not be 0. I came up with the following matrix.

    B(x) = [partial_der_F1 (x1) partial_der_F1 (x3) ; partial_der_F2(x1) partial_der_F2(x3) ]. Evaluating this at 0, I get the following matrix

    [-1 1; 0 1] and the determinant of this matrix is -1 != 0. So I can take the inverse of this matrix

    Thus I can say that for each x2, the following function F(phi1(x2), x2, phi2(x2) ) = 0 where phi1 and phi2 are unique functions.

    I think I have part a right. It is part b I am having problems with

    For part b:

    The formula for the derivative given in class notes is -B(x,g(x))[tex]^{-1}[/tex] A(x,g(x))

    B = matrix calculated above
    A = partial derivative of F1, F2 (2x1 matrix) with respect to x2.

    Now my question is do I have to evaluate this at x2 = 0 or can I just leave it in terms of phi1(x2), x2, phi2(x2).

    Whats confusing me is that if I compute A wrt x2 and evaluate at x2=0, I will get a 0 matrix.

    Thanks in advance

  2. jcsd
  3. May 20, 2008 #2


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    Homework Helper

    I looked over the baby Rudin treatment of this theorem (pg. 224-228) and my conclusion is that the formula given for a derivative in your class notes holds when evaluated at the point. Thus [tex]\left(\phi_1^\prime (0),\phi_2^\prime (0)\right)=(0,0)[/tex] is my conclusion if your calculations are correct.
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