Showing differentiability at 0

[SpringPhysics]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I thought absolute value functions are not differentiable at 0?

 

[Dick]

Homework Statement

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

Homework Equations

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.

 

[SpringPhysics]

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...

 

[Dick]

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.

 

[SpringPhysics]

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.

 

[Dick]

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?

 

[SpringPhysics]

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?

 

[Dick]

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.

 

[SpringPhysics]

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!