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Math Help - Quick implicit differentiation question

  1. #1
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    Quick implicit differentiation question

    The problem is stated as follows:

    ---

    Show that the equation zx^2y^3 + (x+y)e^{z-x}-x^2-x-4y=-3 defines z as a function of x and y near (1,1,1). Also show that (1,1) is a stationary point of this function and determine its kind.

    ---

    The first part is easily shown with the implicit function theorem.

    Differentiating the equation implicitly with respect to x:

    2xy^3z + x^2y^3\frac{\partial z}{\partial x} + e^{z-x} + (\frac{\partial z}{\partial x} -1)(x+y)e^{z-x} -2x -1 = 0, so (x,y,z(x,y)) = (1,1,1) gives 2 + \frac{\partial z}{\partial x} + 1 + 2(\frac{\partial z}{\partial x} - 1)-2-1=0 near (x,y) = (1,1), and consequently, \frac{\partial z}{\partial x} = \frac{2}{3} there.

    Thus, (x,y) = (1,1) isn't a stationary point as the problem claims.

    Feel free to point and laugh at me and tell me that (or preferably why) my arithmetic is wrong.
    Last edited by Combinatus; March 2nd 2011 at 05:05 PM. Reason: None at all.
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  2. #2
    MHF Contributor

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    Looks to me like you are correct. I also get z_x= \frac{2}{3} at (1, 1).
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