Partial derivatives for the sign function

  • Thread starter Telemachus
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  • #1
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Homework Statement


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.
 
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Answers and Replies

  • #2
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All three values are constants. The derivative (or partial derivative) of a constant is zero.
 
  • #4
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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.

Thanks.
 
  • #5
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I think you are making a mistake in your limits. When you evaluate f(x, y), you should get 1, 0, or -1.
 
  • #6
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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.
 
  • #7
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Now I get what you mean. The left part is wrong. I'll keep thinking on it.
 
  • #8
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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.
 
  • #9
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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.
 
  • #10
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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.
 
  • #11
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Thank you Mark.
 

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