# Partial Derivative, piecewise function

1. Oct 22, 2008

### Thomas_

1. The problem statement, all variables and given/known data
Let :
$$f(x,y) = \frac{xy(x^2 - y^2)}{(x^2 + y^2)^2}$$ if $$(x,y) \neq (0,0)$$
$$f(x,y) = 0$$ if $$(x,y) = (0,0)$$

a) Find $$f_{xx}(0,0)$$

b) Find $$f_{xy}(0,0)$$

c) Find $$f_{yx}(0,0)$$

2. Relevant equations
None

3. The attempt at a solution
I'm not sure how to deal with the piecewise defined function, I just took the first and second partial derivatives w/ respect to x (for part a), but then still I can't evaluate the resulting function at (0,0), because at the point it doesn't exist...

2. Oct 22, 2008

### Staff: Mentor

We can't tell whether what you got is correct because you didn't tell us what you got.

3. Oct 23, 2008

### HallsofIvy

Staff Emeritus
That's not really piecewise. f(0, 0) is defined to be 0 in order to make f continuous and so differentiable. You can do the entire problem just ignoring f(0,0).

4. Dec 31, 2008

### MaartenJacobs

I have exactly the same question. The equation is even more or less the same. Mine is:

$$f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$$ if $$(x,y) \neq (0,0)$$
$$f(x,y)=0$$ if $$(x,y)=(0,0)$$.

But I'm still puzzled how to threat the point (0,0). According to my textbook, $$f_{xy}(0,0)\neq f_{yx}(0,0)$$. So, I know more or less the solution, but I'm not sure that my explanation is correct.

Ignoring the point (0,0) I have:

first x, then y:
$$\frac{\partial{f}}{\partial{x}}(x,y)=\frac{-y^5+yx^4+4y^3x^2}{(x^2+y^2)^2}$$

$$\frac{\partial{f}}{\partial{y}\partial{x}}(x,y)=\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}$$

first y, then x:
$$\frac{\partial{f}}{\partial{y}}(x,y)=\frac{x^5-4x^3y^2-y^4x}{(x^2+y^2)^2}$$

$$\frac{\partial{f}}{\partial{x}\partial{y}}(x,y)=\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}$$

So, it is obvious that as long $$x\neq 0$$ and $$y\neq 0$$ that $$\frac{\partial{f}}{\partial{y}\partial{x}}(x,y)=\frac{\partial{f}}{\partial{x}\partial{y}}(x,y)$$.

Now, for (x,y)->(0,0), how do I prove that
$$\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)\neq\frac{\partial{f}}{\partial{x}\partial{y}}(0,0)$$? Clearly, just substituting x and y with 0 results in $$\frac{0}{0}$$.

Is it valid to say the following:
For$$\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)$$, first let x-> 0 first, resulting in $$\frac{\partial{f}}{\partial{y}\partial{x}}(0,y)=\frac{-y^6}{y^6}=-1, \forall y$$ and thus$$\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)=-1$$?

For$$\frac{\partial{f}}{\partial{x}\partial{y}}(0,0)$$, first let y-> 0 first, resulting in $$\frac{\partial{f}}{\partial{x}\partial{y}}(x,0)=\frac{x^6}{x^6}=1, \forall x$$ and thus$$\frac{\partial{f}}{\partial{x}\partial{y}}(0,0)=1$$?

This is more or less a guess. I have no clue whether it is correct and if it is correct how to justify this. Especially, I have no explanation why for $$\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)$$, I first let x->0 and not y->0.

Any help on this would be most welcome and will be highly appreciated.

(and happy new year to you all)

5. Dec 31, 2008

### HallsofIvy

Staff Emeritus
The partial derivatives are defined by the limits
$$f_x(x_0,y_0)= \lim_{h\rightarrow 0} \frac{f(x_0+h, y_0)- f(x_0,y_0)}{h}$$
and
$$f_y(x_0,y_0)= \lim_{h\rightarrow 0} \frac{f(x_0,y_0+h)- f(x_0,y_0)}{h}$$
Here, $(x_0,y_0)= (0,0)$ and with h non-zero, in calculating the limit, we are not at (0,0) and so use the other formula.

Here,
[tex]f_x(0,0)= \lim_{h\rightarrow 0} \frac{\frac{(x+h)(0)((x+h)^2- 0^2)}{(x+h)^2- 0^2)- 0}}{h}= \lim_{h\rightarrow 0} 0/h= 0[/itex]
[tex]f_y(0,0)= \lim_{h\rightarrow 0} \frac{\frac{(0)(y+h)(0^2- (y+h)^2)}{0^2-(y+h)^2- 0}}{h}= \lim_{h\rightarrow 0} 0/h= 0[/itex]
While $f_x$ and $f_y$ for (x,y) not equal to (0,0) is given by differentiating f in the usual way.

Once you know $f_x$ and $f_y$ for (x,y) close to (0,0) do exactly the same thing with this new function to find the second derivative.