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Continuous Partial Derivative

  1. Feb 4, 2009 #1
    If [tex]u : R^2 \to R[/tex] has continuous partial derivatives at a point [tex](x_0,y_0)[/tex] show that:

    [tex]u(x_0+\Delta x, y_0+\Delta y) = u_x(x_0,y_0) + u_y(x_0,y_0) + \epsilon_1 \Delta x + \epsilon_2 \Delta y[/tex], with [tex]\epsilon_1,\, \epsilon_2 \to 0[/tex] as [tex]\Delta x,\, \Delta y \to 0[/tex]

    I know this can be proved using MVT, but I tried to prove this another way only my proof doesn't use the continuity of both partial derivatives so I thought there'd be an error but I couldn't spot it, so I was hoping somebody else could. Here is my proof:

    I will use the result that for a differentiable function f,

    [tex]f(x+h) = f(x) + f'(x) h + \epsilon h[/tex] where [tex]\epsilon[/tex] is a function of h and goes to 0 as h goes to zero. (Follows from the definition of the derivative.)

    [tex]u(x_0+\Delta x,y_0+\Delta y) - u(x_0,y_0)[/tex]

    [tex]= u(x_0,y_0+\Delta y) + u_x(x_0,y_0+\Delta y) \Delta x + \epsilon_1 \Delta x - u(x_0,y_0)[/tex]

    [tex]= (u(x_0,y_0) + u_y(x_0,y_0) \Delta y + \epsilon_2\Delta y) + u_x(x_0,y_0+\Delta y) \Delta x + \epsilon_1\Delta x - u(x_0,y_0)[/tex]

    [tex]= u_y(x_0,y_0) \Delta y + u_x(x_0,y_0+\Delta y) \Delta x + \epsilon_1 \Delta x + \epsilon_2 \Delta y[/tex]

    Here [tex]\epsilon_1,\, \epsilon_2 \to 0[/tex] as [tex]\Delta x,\, \Delta y \to 0[/tex].

    I will be done if I can prove [tex]u_x(x_0, y_0 + \Delta y) \Delta x = u_x(x_0, y_0)\Delta x + \epsilon_3 \Delta x[/tex] with [tex]\epsilon_3 \to 0[/tex] as [tex]\Delta x,\, \Delta y \to 0[/tex], but this follows from the continuity of u_x at the point [tex](x_0, y_0)[/tex].

  2. jcsd
  3. Feb 5, 2009 #2
    I'm not sure if this is the cause of your problem, but you seem to be disregarding the fact that [itex]\epsilon_1 \equiv \epsilon_1(y_0 + \Delta y), \epsilon_2 \equiv \epsilon_2(x_0 + \Delta x)[/itex]. It doesn't follow from the fact that for every Delta y, epsilon 1 goes to zero if Delta x goes to zero that epsilon 1 goes to zero if we approach the point from any direction.
  4. Feb 5, 2009 #3
    hi! could you please clear this doubt...
    there is a theorem stating that if the first partial derivatives are continuous, then the function in 2 variables is differentiable. Is it enough to prove this in case of a split function?
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