Definition of multivariable function classes doubt

Question

I was given this example by my teacher:

$f(x)=\begin{cases} 0 & (x,y) = 0 \\ \frac{x^2y^2}{x^4 + y^2} & (x,y)\neq (0,0)\ \end{cases} $

He proved diferentiablity for all $\mathbb{R}^2$.

Then he went to prove wether the function was class $C^1$ or not. He used $y=x^2$ to prove it wasn't. And that's what put me in doubt.

Doesn't diferentiability in $\mathbb{R}^2$ imply it's partial derivatives are continuous, therefore $C^1$. Am I missing something in this example?

Continuity and differentiability are checked in this answer to another question, but this is not a duplicate of that question. — Mark S., Mar 24 '21 at 09:40

José Carlos Santos · Accepted Answer · 2021-03-24T18:47:39.747

0

No, differentiability does not imply that the partial derivatives are continuous (it's the other way around). Take, for instance$$f(x,y)=\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ otherwise.}\end{cases}$$

edited Mar 24 '21 at 18:47

answered Mar 24 '21 at 09:39

José Carlos Santos

427,504

Thanks, so to check if a function is class $C^1$ I have only to worry about continuity of partial derivatives? – José Pedro Ferreira Mar 24 '21 at 09:48
Yes, that is correct. – José Carlos Santos Mar 24 '21 at 10:03

Definition of multivariable function classes doubt

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