Why is f(x,y) not differentiable at (0,0)?

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SUMMARY

The function f(x,y) = sqrt(abs(xy)) is not differentiable at the point (0,0) due to the behavior of its directional derivatives. While the function is continuous at (0,0), the limit of the directional derivatives does not converge uniformly, leading to the conclusion that the gradient does not exist at this point. Specifically, approaching (0,0) along different paths yields different limits, confirming the lack of differentiability. The analysis reveals that the limit along the 45-degree line results in a value of 1, contrasting with the expected zero gradient.

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wakko101
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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.
 
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Show that's it's not continuous?
 
Try taking limits and show that the limit doesn't exist at (0,0).
 
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?
 
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.
 
Since this problem is old now, I will give out my full solution now:

http://img113.imageshack.us/img113/5482/mysolutiontf9.jpg

JFonseka said:
Show that's it's not continuous?
Actually, f(x,y) is continuous at 0:

http://img404.imageshack.us/img404/1144/continuityproofqf0.jpg
 
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