Vector calculus - Prove a function is not differentiable at (0,0)

In summary, the conversation discusses the differentiability of the function ##f(x)## at ##(0,0)##. While it is shown that the function has partial derivatives and is continuous at this point, it is also proven that it is not differentiable. There was initially some confusion about the proof, but it was eventually understood that the use of vector coordinates was necessary to show the lack of differentiability.
  • #1
104
4
##f\left(x\right)=\begin{cases}\sqrt{\left|xy\right|}sin\left(\frac{1}{xy}\right)&xy\ne 0\\ 0&xy=0\end{cases}##
I showed it partial derivatives exist at ##(0,0)##, also it is continuous as ##(0,0)##
but now I have to show if it differentiable or not at ##(0,0)##.
According to answers it is not and they proved by showing the vector coordinates at ##(1,1)## does not have a limit.
But I dont want a proof like that, I tried using definition and got stuck...
I know my Linear transformation is basically 0. but still got in trouble of the definition
 
Physics news on Phys.org
  • #2
Hi, is not needed anymore, at the end I managed to understand why they did vector coordinates, it is pratically trivial as I see it now
Thank thought :)
 

Suggested for: Vector calculus - Prove a function is not differentiable at (0,0)

Replies
5
Views
2K
Replies
3
Views
892
Replies
5
Views
1K
Replies
6
Views
962
Replies
11
Views
976
Replies
0
Views
2K
Replies
12
Views
982
Replies
6
Views
288
Back
Top