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Differentiable Vector Functions

  1. Mar 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Show through direct application of definition that the function

    [tex]f(x,y) = xy[/tex]

    is differentiable at (1,1)

    3. The attempt at a solution

    I know that all functions of the class C1 are differentiable and that a function is of the class C1 if its partial derivatives exists and are continuous.

    So do I prove that the two partial derivatives exist by showing that the following to limits exists?

    [tex]\frac{\partial f}{\partial x}(a,b) = \lim_{h \to 0} \frac{f(a+h,b) - f(a,b)}{h}[/tex]

    [tex]\frac{\partial f}{\partial y}(a,b) = \lim_{k \to 0} \frac{f(a,b+k) - f(a,b)}{k}[/tex]

    The, the following obtains:

    [tex]\frac{\partial f}{\partial x}(1,1) = \lim_{h \to 0} \frac{f(1+h,1) - f(1,1)}{h} = \lim_{h \to 0} \frac{(1+h)1 - 1}{h} = \lim_{h \to 0}\frac{h}{h} = 1[/tex]

    [tex]\frac{\partial f}{\partial y}(1,1) = \lim_{k \to 0} \frac{f(1,1+k) - f(1,1)}{k} = \lim_{k \to 0} \frac{(1(1+k) - 1}{k} = \lim_{k \to 0}\frac{k}{k} = 1[/tex]

    Since both partial derivatives are elementary functions, they are continuous, correct? If so, this ought to show that the function is differentiable in (1,1)?

    Thank you for your time. Have a nice day.
  2. jcsd
  3. Mar 1, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, given that you have the theorem "if the partial derivatives of function f exist and are continuous at a point then the function is differentiable there" then what showing the partial derivatives are continuous would do it. However, I don't see that you have proven the partial derivatives are continuous. You have only calculated the partial derivatives at (1, 1). That does not show that they are elementary functions in a neighborhood of (1, 1) or that they are continuous.

    One could also show it, without that theorem, directly from the definition of "differentiable".
  4. Mar 1, 2008 #3
    So I have shown that the partial derivatives exist at the point (1,1). Can I then make the following argument:

    1. If the functions have partial derivatives at all, they must be [insert the two partial derivatives here] (by taking the two partial derivative of the function with the short-hand rules)
    2. If the partial derivatives exist, they are elementary functions and thus continuous. (by demonstration)
    3. The partial derivatives exists at the point (1,1)
    4. The partial derivatives are continuous at point (1,1) (from 2).
    5. The function is differentiable at point (1,1) (from 3&4)

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