1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Partial derivatives for the sign function

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi there. Well, I've got some doubts on the partial derivatives for the next function:

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

    So, what I got is:

    [tex]f(x,y)=f(x)=\begin{Bmatrix}{ 1}&\mbox{ if }& (y-x^2)(y-2x^2)>0\\0 & \mbox{if}& (y-x^2)(y-2x^2)=0\\-1 & \mbox{if}& (y-x^2)(y-2x^2)<0\end{matrix}[/tex]

    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.
    Last edited: Sep 19, 2010
  2. jcsd
  3. Sep 19, 2010 #2


    Staff: Mentor

    All three values are constants. The derivative (or partial derivative) of a constant is zero.
  4. Sep 19, 2010 #3
  5. Sep 20, 2010 #4
    This is the way I've solved it for x. All three values are constants, but I cant just evaluate at (0,0) because its a closed set at that point.

    If [tex]x=y=0[/tex]
    [tex]\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[/tex]

    If [tex]y=x^2[/tex], [tex](x,y)=(x_0,x_0^2)[/tex]

    [tex]\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}}=[/tex]

    [tex]=\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}[/tex]

    If [tex](x,y)=(x_0,2x_0^2)[/tex]

    [tex]\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}}=[/tex]

    [tex]=\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[/tex]

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

    Is this right?

    For y I've proceeded same way. I think its the right way for solving it.

  6. Sep 20, 2010 #5


    Staff: Mentor

    I think you are making a mistake in your limits. When you evaluate f(x, y), you should get 1, 0, or -1.
  7. Sep 20, 2010 #6
    Yes, but what about the derivatives? I've received an email with a message from you that doesn't appear here. I think that post pointed the right direction.

    I'm not sure about what to do with the boundary points.
  8. Sep 20, 2010 #7
    Now I get what you mean. The left part is wrong. I'll keep thinking on it.
  9. Sep 20, 2010 #8


    Staff: Mentor

    I posted something, then deleted, so that might be the email you received.

    I think it's helpful to look at the plane as three regions:
    Region I - the points above the curve y = 2x^2
    Region II - the points below y = 2x^2 and above y = x^.
    Region III - the points below y = x^2.

    These regions do not include the boundary curves.
    In regions I and III, f(x, y) = 1. In region II, f(x, y) = -1.

    Clearly, for every point in any of these regions, both partials are zero.

    On the boundary curves, f(x, y) = 0.
  10. Sep 20, 2010 #9
    How do I demonstrate it using the limit definition? I see I'm committing a mistake when I try, but I don't know how to do it right.
  11. Sep 20, 2010 #10


    Staff: Mentor

    I think you need to do one-sided limits (h > 0 and h < 0), and also make different cases for when x < 0 and x > 0.

    For example, if (x0, y0) is a point on the graph of y = 2x^2, (x0 + h, y0) will be a point in region II if h > 0 and x0 > 0, but will be a point in region I if h > 0 and x0 < 0.

    Draw a graph of the three regions and boundary curves and look at different possibilities. That's what I would advise.
  12. Sep 20, 2010 #11
    Thank you Mark.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook