# Partial derivatives for the sign function

• Telemachus
In summary, Homework Statement:The user has 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.The user has found that all three values are constants, but can't just evaluate at (0,0) because its a closed set at that point.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
Telemachus

## Homework Statement

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:

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

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:
All three values are constants. The derivative (or partial derivative) of a constant is zero.

Thanks!

This is the way I've solved it for x. All three values are constants, but I can't just evaluate at (0,0) because its a closed set at that point.

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?

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

Thanks.

I think you are making a mistake in your limits. When you evaluate f(x, y), you should get 1, 0, or -1.

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.

Now I get what you mean. The left part is wrong. I'll keep thinking on it.

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.

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.

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.

Thank you Mark.

## 1. What is the sign function?

The sign function, also known as the signum function, is a mathematical function that returns the sign of a given number. It is commonly denoted as "sgn(x)" and has a value of -1, 0, or 1 depending on whether the input is negative, zero, or positive, respectively.

## 2. What are partial derivatives?

Partial derivatives are a type of derivative that measures the rate of change of a function with respect to one of its variables while holding all other variables constant. In other words, it calculates how much a function changes in response to a small change in one of its inputs.

## 3. How are partial derivatives calculated for the sign function?

For the sign function, the partial derivative is calculated by taking the limit as the change in the input variable approaches zero. This results in a derivative of either -1 or 1, depending on the sign of the input variable.

## 4. What are the applications of partial derivatives for the sign function?

Partial derivatives for the sign function are commonly used in optimization problems, such as finding the maximum or minimum of a function. They are also useful in economics and finance for analyzing the sensitivity of a variable to changes in other variables.

## 5. Are there any limitations to using partial derivatives for the sign function?

Yes, there are limitations to using partial derivatives for the sign function. Since the sign function is discontinuous at x=0, the derivative is undefined at this point. Additionally, the sign function is not differentiable at x=0, which means that the derivative does not exist at this point.

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