Results 1 to 3 of 3

Math Help - Partial derivatives for the sign function

  1. #1
    Member
    Joined
    May 2010
    Posts
    241

    Partial derivatives for the sign function

    Hi there. Well, I've got some doubts on the partial derivatives for the next function:

    f(x,y)=sg\{(y-x^2)(y-2x^2)\} Where sg is the sign function.

    So, what I got is:

    f(x,y)=f(x)=\begin{Bmatrix}{ 1}&\mbox{ si }& (y-x^2)(y-2x^2)>0\\0 & \mbox{si}& (y-x^2)(y-2x^2)=0\\-1 & \mbox{si}& (y-x^2)(y-2x^2)<0\end{matrix}

    How should I get the partial derivatives? I'm sure that for the zero I must use the definition. But in the other cases should I?

    Bye, and thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Because f(*,*) is a constant in the regions where is (y-x^{2})\ (y-2\ x^{2}) >0 or (y-x^{2})\ (y-2\ x^{2}) < 0 , here the partial derivatives are both equal to 0. Where is (y-x^{2})\ (y-2\ x^{2}) = 0 the partial derivatives don't exist...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2010
    Posts
    241
    How did you solve it?

    Is this the right way?


    If x=y=0
    \displaystyle\frac{\partial f}{\partial x}=\displaystyle\lim_{x \to{0}}{\displaystyle\frac{f(x,0)-f(0,0}{x}}=\displaystyle\lim_{x \to{0}}{\displaystyle\frac{(0-x^2)(0-2x^2-0)}{x}}=\displaystyle\lim_{x \to{0}}{\displaystyle\frac{2x^4}{x}}=0

    If y=x^2, (x,y)=(x_0,x_0^2)

    \displaystyle\frac{\partial f}{\partial x}=\displaystyle\lim_{h \to{0}}{\displaystyle\frac{f(x_0+h,x_0^2)-f(x_0,x_0^2)}{h}}=\displaystyle\lim_{h \to{0}}{\displaystyle\frac{(x_0^2-(x_0+h)^2)(x_0^2-2(x_0+h)^2)-0}{h}}=\displaystyle\lim_{h \to{0}}{\displaystyle\frac{(x_0^2-x_0^2-2x_0h-h^2)(x_0^2-2x_0^2-4x_0h-h^2)}{h}}=

    =\displaystyle\lim_{h \to{0}}{\displaystyle\frac{-h(2x_0-h)(x_0^2+4x_0h+h^2)}{h}}=\displaystyle\lim_{h \to{0}}{-(2x_0-h)(x_0^2+tx_0h+h^2)=-2x_0^3}

    If (x,y)=(x_0,2x_0^2)

    \displaystyle\frac{\partial f}{\partial x}=\displaystyle\lim_{h \to{0}}{\displaystyle\frac{f(x_0+h,2x_0^2)-f(x_0,2x_0^2)}{h}}=\displaystyle\lim_{h \to{0}}{\displaystyle\frac{(2x_0^2-(x_0+h)^2)(2x_0^2-2(x_0+h)^2)-0}{h}}=\displaystyle\lim_{h \to{0}}{\displaystyle\frac{(2x_0^2-x_0^2-2x_0h-h^2)(2x_0^2-2x_0^2-4x_0h-h^2)}{h}}=

    =\displaystyle\lim_{h \to{0}}{\displaystyle\frac{(x_0^2-2x_0h-h^2)(-h(4x_0+h))}{h}}=\displaystyle\lim_{h \to{0}}{-(x_0^2-2x_0h-h^2)(4x_0+h)=-4x_0^2

    As -4x_0^2\neq{-2x_0^3} the derivative doesn't exist at the point (0,0).

    Is this right?

    Thanks.
    Last edited by Ulysses; September 20th 2010 at 08:23 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Production Function (Partial Derivatives)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 15th 2012, 02:24 AM
  2. Find the partial derivatives of the function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 15th 2010, 07:48 PM
  3. Partial derivatives of implicit function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 22nd 2010, 11:58 AM
  4. Increasing function, partial derivatives
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 13th 2009, 08:23 PM
  5. Partial derivatives- production function
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 17th 2008, 01:38 PM

Search Tags


/mathhelpforum @mathhelpforum