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

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SUMMARY

The function defined as \( f(x) = \begin{cases} \sqrt{|xy|} \sin\left(\frac{1}{xy}\right) & xy \ne 0 \\ 0 & xy = 0 \end{cases} \) is shown to have existing partial derivatives and continuity at the point (0,0). However, it is proven not to be differentiable at (0,0) by demonstrating that the limit of the vector coordinates at (1,1) does not exist. The discussion emphasizes the importance of understanding the definition of differentiability and the application of vector coordinates in proving non-differentiability.

PREREQUISITES
  • Understanding of partial derivatives and continuity in multivariable calculus
  • Familiarity with the definition of differentiability in the context of vector calculus
  • Knowledge of limits and their properties in two dimensions
  • Basic concepts of linear transformations in calculus
NEXT STEPS
  • Study the definition of differentiability in multivariable calculus
  • Learn about limits in vector calculus, particularly in two dimensions
  • Explore examples of functions that are continuous but not differentiable
  • Investigate the application of linear transformations in proving differentiability
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and vector calculus, as well as educators looking to deepen their understanding of differentiability concepts.

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##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
 
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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 :)
 

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