Results 1 to 5 of 5

Math Help - deferentiability question ts7 q1

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    deferentiability question ts7 q1

    i have this function
    f(x,y)=\begin{cases}x^{2}+y^{2}ln(x^{2}+y^{2}) & (x,y)\neq(0,0)\end{cases}
    and f(x,y)=0 on point (0,0)

    i dont know the proper english term
    i need to prove that f(x,y) is deferentialbe

    so by the definition
    f(x,y)-f(0,0)-f_{x}(x-0)-f_{y}(y-0)=\epsilon_{1}(x-0)+\epsilon_{2}(y-0)
    and i need to find those epsilons which will make the sude to be equal
    and both epsilons will go to zeroo when x and y are going to zero


    i dont know what to put instead of "f" in each term
    because the function is stplitted
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: deferentiability question ts7 q1

    Quote Originally Posted by transgalactic View Post
    i have this function
    f(x,y)=\begin{cases}x^{2}+y^{2}ln(x^{2}+y^{2}) & (x,y)\neq(0,0)\end{cases}
    and f(x,y)=0 on point (0,0)

    i dont know the proper english term
    i need to prove that f(x,y) is deferentialbe

    so by the definition
    f(x,y)-f(0,0)-f_{x}(x-0)-f_{y}(y-0)=\epsilon_{1}(x-0)+\epsilon_{2}(y-0)
    and i need to find those epsilons which will make the sude to be equal
    and both epsilons will go to zeroo when x and y are going to zero


    i dont know what to put instead of "f" in each term
    because the function is stplitted

    What you did? I really don't understand...

    First check if f(x,y) is continuous in point (0,0). If it isn't therefor you can conclude that f(x,y) is not deferentiable.

    Edit:
    Are you sure about the function, perhaps you meant to write: (x^2+y^2)ln(x^2+y^2) ?
    Last edited by Also sprach Zarathustra; June 16th 2011 at 02:38 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    Re: deferentiability question ts7 q1

    this is a question of checking by definition
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: deferentiability question ts7 q1

    Quote Originally Posted by transgalactic View Post
    this is a question of checking by definition

    Perhaps the function is

    f(x,y)=\begin{Bmatrix} (x^2+y^2)\ln (x^2+y^2) & \mbox{ if }& (x,y)\neq (0,0)\\0 & \mbox{if}& (x,y)=0\end{matrix}

    Then, \frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=\ldots=0 so, if f is differentiable at (0,0) the only possible differential is \lambda:\mathbb{R}^2\to \mathbb{R},\;\lambda (h)=0 . Now you have to study the limit

    \lim_{h \to 0}{\frac{|f(h)-f(0)-\lambda(h)|}{ \left\|{h}\right\|}}=\lim_{h \to 0}{\frac{|f(h)|}{ \left\|{h}\right\|}}

    The function is differentiable at (0,0) iff the limit is 0 .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    Re: deferentiability question ts7 q1

    i need to use only the epsilon method

    anybody knows it?
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum