# Thread: Second partial derivative of this function at (0,0)

1. ## Second partial derivative of this function at (0,0)

Let $\displaystyle f: \mathbb {R}^2 \rightarrow \mathbb {R}$ be defined by $\displaystyle f(x,y)= \frac {xy(x^2-y^2)}{x^2+y^2}$ if $\displaystyle (x,y) \neq (0,0)$ and $\displaystyle f(0,0)=0$

Show that $\displaystyle \frac { \partial ^2 f }{ \partial x \partial y }$ and $\displaystyle \frac { \partial ^2 f }{ \partial y \partial x }$ exist at (0,0) but not equal.

Now, should I go ahead and take the partial derivative as how it looks then plug in (0,0)? Or should I approach this with the limit?

I just want to know if I'm going to start this one correctly. Thank you!

2. Check out this thread for a similar example. The answer to the question is that at points where the defining formula for the function breaks down (as it does at the origin in this example) you have to go back to the limit definition of derivative.

3. I actually have the same problem.

I am following along with the link, and I thought that I had the right answer, but something seems strange.

So using the limit definition I"m getting that df/dx(0,y)= lim(h->0) of (hy)(h^2-y^2)/(h^2+y^2)= -hy which gives that df/dx (0,0)=0 so it's works. But with the second partial with respect to y I have.

d^2f/dxdy(0,0) = lim (k->0) of -hk-0/k = -h. But that doesn't give me that d^2f/dxdy(0,0)=0?

Maybe I'm doing something wrong?

4. Originally Posted by EricaMae
I actually have the same problem.

I am following along with the link, and I thought that I had the right answer, but something seems strange.

So using the limit definition I"m getting that df/dx(0,y)= lim(h->0) of (hy)(h^2-y^2)/(h^2+y^2)= -hy which gives that df/dx (0,0)=0 so it's works. But with the second partial with respect to y I have.

d^2f/dxdy(0,0) = lim (k->0) of -hk-0/k = -h. But that doesn't give me that d^2f/dxdy(0,0)=0?

Maybe I'm doing something wrong?
$\displaystyle \frac{\partial f}{\partial x}(0,y) = \lim_{h\to0}\frac{f(h,y)-f(0,y)}h = \lim_{h\to0}\frac{hy(h^2-y^2)}{{\color{red}h}(h^2+y^2)} = -y$.