# Derivative of absolute value

1. Nov 6, 2007

### wakko101

Hello,

My question is as follows: Show that the function f(x,y) = sqrt(abs(xy)) is not differentiable at (0,0).

I was going to go with trying to show that the directional derivatives don't all exist here, but that would require finding the gradient, and I always get confused when trying to take the derivative of an absolute value. Essentially, this means that for xy larger than 0, f = sqrt(xy) and for xy smaller than 0, f = sqrt(-xy). But, of course, you can't have the square root of a negative number, so I'm confused....what should I do?

Thanks,
W.

2. Nov 6, 2007

### JFonseka

Show that's it's not continuous?

3. Nov 6, 2007

### andytoh

Try taking limits and show that the limit doesn't exist at (0,0).

4. Nov 6, 2007

### wakko101

but which limits? If I want to use the definition of a derivative for multi variables, then I still need the gradient, don't I?

5. Nov 6, 2007

### andytoh

Use the limit definition of Df(x,y). In this case, if f were differentiable at 0, then Df(0,0) would be the zero map. On the other hand, by approaching zero along the 45 degree line in the first quadrant one would then have the limit to be 0 in spite of the fact that the limit is clearly 1. You'll have to try it to see what I mean.

6. Nov 11, 2007

### andytoh

Since this problem is old now, I will give out my full solution now:

http://img113.imageshack.us/img113/5482/mysolutiontf9.jpg [Broken]

Actually, f(x,y) is continuous at 0:

http://img404.imageshack.us/img404/1144/continuityproofqf0.jpg [Broken]

Last edited by a moderator: May 3, 2017