# Showing differentiability at 0

• SpringPhysics
In summary: Oh wow, I feel so stupid. I forgot that I was differentiating at 0, so I had point a = (x,y) and I was stressing out about why the limit wouldn't go to 0. LOL. Thank you greatly for your help and patience!
SpringPhysics

## Homework Statement

Show that f(x,y) = |xy| is differentiable at 0.

## The Attempt at a Solution

I thought absolute value functions are not differentiable at 0?

SpringPhysics said:

## Homework Statement

Show that f(x,y) = |xy| is differentiable at 0.

## The Attempt at a Solution

I thought absolute value functions are not differentiable at 0?

f(x)=|x| isn't differentiable at x=0. f(x,y)=|xy| at x=0 and y=0 is. Now try and show it.

I am attempting to find a matrix A so that the definition of differentiability (in higher dimensions) holds...but I cannot seem to do so because I cannot cancel out the absolute value of h depending on whether h is negative or positive. Is this the right direction?

Sorry, I am not even sure which definition/theorem of differentiability I should use, since I have only learned of the definition and the C1 condition, but this function isn't even C1.

EDIT: Do I use the sgn(x) function? f(x,y) = |xy| = |x||y|. So the partial derivatives are x|y|/|x| and y|x|/|y|, unless x, y are zero, in which case the partial derivatives are 0. So then the partial derivatives exist. But they are still not continuous at 0...

Last edited:
SpringPhysics said:
I am attempting to find a matrix A so that the definition of differentiability (in higher dimensions) holds...but I cannot seem to do so because I cannot cancel out the absolute value of h depending on whether h is negative or positive. Is this the right direction?

Sorry, I am not even sure which definition/theorem of differentiability I should use, since I have only learned of the definition and the C1 condition, but this function isn't even C1.

EDIT: Do I use the sgn(x) function? f(x,y) = |xy| = |x||y|. So the partial derivatives are x|y|/|x| and y|x|/|y|, unless x, y are zero, in which case the partial derivatives are 0. So then the partial derivatives exist. But they are still not continuous at 0...

Yes, I don't think the partial derivatives are continuous at (0,0). But that shouldn't stop you. Try to show the matrix A=0 works. Try thinking about polar coordinates.

I had previously attempted A = 0. Now that I use polar coordinates, I end up with basically the same problem...except that my x and y are now r cos theta and r sin theta. I know that each are bounded by r, but that still leaves me to deal with the r. Everything else goes to zero except the terms with the constant r. Am I on the right track? Thanks so much for your help.

SpringPhysics said:
I had previously attempted A = 0. Now that I use polar coordinates, I end up with basically the same problem...except that my x and y are now r cos theta and r sin theta. I know that each are bounded by r, but that still leaves me to deal with the r. Everything else goes to zero except the terms with the constant r. Am I on the right track? Thanks so much for your help.

Well, you need to show lim |xy|/sqrt(x^2+y^2)=0 as r->0, right? Wouldn't that show A=0 is the matrix derivative?

Dick said:
Well, you need to show lim |xy|/sqrt(x^2+y^2)=0 as r->0, right? Wouldn't that show A=0 is the matrix derivative?

Sorry, I just lost you. I thought the definition for the derivative is
lim [f(a+h) - f(a) - Ah]/|h| as h -> 0, for h in R2?

SpringPhysics said:
Sorry, I just lost you. I thought the definition for the derivative is
lim [f(a+h) - f(a) - Ah]/|h| as h -> 0, for h in R2?

The definition of A is lim [f(a+h) - f(a) - Ah]/|h|=0 as h -> 0, for h in R^2
a=(0,0). h=(x,y). f(a)=|0*0|=0. Ah=0 since A=0. f(a+h)=|xy|. |h|=sqrt(x^2+y^2). Try and stay with me, buddy.

Dick said:
a=(0,0). h=(x,y). f(a)=|0*0|=0. Ah=0 since A=0. f(a+h)=|xy|. |h|=sqrt(x^2+y^2). Try and stay with me, buddy.

Oh wow, I feel so stupid. I forgot that I was differentiating at 0, so I had point a = (x,y) and I was stressing out about why the limit wouldn't go to 0. LOL. Thank you greatly for your help and patience!

## 1. What is differentiability at 0?

Differentiability at 0 refers to a mathematical property of a function where its derivative exists at the point x=0. This means that the function has a well-defined tangent line at x=0 and can be approximated by a linear function in the neighborhood of x=0.

## 2. How do you show differentiability at 0?

To show differentiability at 0, you must first determine if the function is continuous at x=0. If it is, then you can take the limit of the difference quotient as x approaches 0 to find the derivative at 0. If the limit exists, then the function is differentiable at 0.

## 3. Why is differentiability at 0 important?

Differentiability at 0 is an important concept in calculus and real analysis because it allows us to approximate nonlinear functions with linear ones, making it easier to analyze and solve problems. It also has many applications in physics, engineering, and other fields.

## 4. Can a function be differentiable at 0 but not continuous?

No, a function cannot be differentiable at 0 if it is not continuous at that point. This is because differentiability requires continuity, but continuity does not necessarily imply differentiability.

## 5. How does differentiability at 0 relate to the differentiability of a function as a whole?

If a function is differentiable at 0, it does not necessarily mean that it is differentiable at all other points. However, if a function is differentiable at all points, it is also differentiable at 0. In other words, differentiability at 0 is a necessary but not sufficient condition for the differentiability of a function as a whole.

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