# Is this function differentiable?

1. Oct 11, 2011

### _Steve_

Hey guys, I'm just wondering if I got this question right:

Discuss where the following in $R^{2}$ is differentiable:
$f(x,y)=(x^{3}+y^{3})^{2/3}$

So I take the partial derivative:

$f_{x}(x,y)=\frac{2x^{2}}{(x^{3}+y^{3})^{1/3}}$

and see that f(x,y) might not be differentiable at (0,0), so I used the definition of a partial derivative to see if the partial derivatives exist at (0,0):

$f_{x}(0,0)=lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}=0$

I thought this was the end of the question, since it shows that the partial derivative at (0,0) does exist, and thus it's continuous for all (x,y) (showing that f(x,y) is differentiable)
But under the question there's a hint, it says:

Hint: Verify the handy inequality $|x^{3}+y^{3}| \leq 2(x^{2}+y^{2})^{3/2}$

Did I miss something?

2. Oct 11, 2011

### SammyS

Staff Emeritus
Look at f(x, y) in the neighborhood of the line y=-x .

Look at fx(x, y) in the neighborhood of the line y=-x .

Last edited: Oct 11, 2011
3. Oct 11, 2011

### _Steve_

So I'm not done the question? What am I looking for at the moment to show that it's differentiable?

Sorry, just trying to understand the question better! :)

4. Oct 11, 2011

### _Steve_

Ohh, I see what you're saying, when y=-x, the denominator of fx = 0, so would I just say that f(x,y) isn't differentiable when y=-x? Or do I have to go back to limits and use squeeze theorem somehow? (Just a cause cause the hint looks like squeeze theorem :P)