# Thread: Is this function differentiable?

1. ## Is this function differentiable?

$f(x,y)=\frac{x^{4}}{x^{4}+y^{4}}$ if $(x,y)\ne(0,0)$

$f(x,y)=0$ if $(x,y)=(0,0)$

I have no idea how to start it, thank you very much!

2. Originally Posted by Bop
$f(x,y)=\frac{x^{4}}{x^{4}+y^{4}}$ if $(x,y)\ne(0,0)$

$f(x,y)=0$ if $(x,y)=(0,0)$

I have no idea how to start it, thank you very much!
You would start by trying to differentiate it! I assume you know that a function of two variables is differentiable at a point if its partial derivatives exist and are continuous in some neighborhood of that point.

It should be clear that there is no problem anywhere except possibly at (0,0). What are the partial derivatives? Are they continuous at (0,0)?

3. There is no problem when $x^{2}=-y^{2}$?

To be able to differentiate it must be defined in the point and be continuous, it is defined so we have to watch the limits:

When $x=0 \rightarrow f(x,y)=\frac{0}{y^{4}}=0$

When $y=0 \rightarrow f(x,y)=\frac{x^{4}}{x^{4}}=1$

As $0\ne1$ there is no limit, so the function is no differentiable in (0,0). Right?

4. Or is it necessary calculate partial derivatives?

5. Originally Posted by Bop
There is no problem when $x^{2}=-y^{2}$?l
x and y are both real numbers. Theirs squares are both non-negative so $x^2$ can't be equal to $-y^2$!

To be able to differentiate it must be defined in the point and be continuous, it is defined so we have to watch the limits:

When $x=0 \rightarrow f(x,y)=\frac{0}{y^{4}}=0$

When $y=0 \rightarrow f(x,y)=\frac{x^{4}}{x^{4}}=1$

As $0\ne1$ there is no limit, so the function is no differentiable in (0,0). Right?[/QUOTE]
Yes, excellent point!

6. x and y are both real numbers. Theirs squares are both non-negative so can't be equal to !
Yes, I didn't realize it! Thank you!

If there were limit in (0,0) it would be yet necessary calculate partial derivatives?