# Math Help - Evaluating Partial Derivative

1. ## Evaluating Partial Derivative

The question is:
Evaluate $f_{xy}(0, 0)$ and $f_{yx}(0, 0)$ for $f(x, y) = xy\frac{x^2−y^2}{x^2+y^2}$.
What would you say about Clairaut's theorem?

I found fxy(x,y) but then I realized... wouldn't this be undefined if you evaluate at f(0,0)?

Thanks.

2. If you read the text of Clairaut's theorem carefully, you might see what the problem is getting at. What did you get for evaluating the mixed partial derivatives at the origin?

3. Originally Posted by Ackbeet
If you read the text of Clairaut's theorem carefully, you might see what the problem is getting at. What did you get for evaluating the mixed partial derivatives at the origin?
For $f_{xy}=\frac{x^8-y^8}{(x^2+y^2)^4}$ Is it not undefined if evaluated at the origin? Does this mean that the theorem doesn't apply?

4. Hmm. Not sure you took the derivative correctly. Just to be clear, the function is defined by

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

Is that correct? If so, what do you get, precisely, for $f_{x}(x,y)$? And then what do you get for $f_{xy}(x,y)$? Please show your steps.

5. $f(x,y) = \frac{x^3y - xy^3}{x^2+y^2}$

$f_x(x,y) = \frac{(3x^2y-y^3)(x^2+y^2) - 2x(x^3y-xy^3)}{(x^2+y^2)^2}
= \frac{x^4y+4x^2y^3-y^5}{(x^2+y^2)^2}$

$f_{xy}(x,y) = \frac{(x^4+12y^2x^2-5y^4)(x^2+y^2)^2 - 2(x^2+y^2)(2y(x^4y+4x^2y^3-y^5)}{(x^2+y^2)^4}$
$= \frac{(x^8+14x^6y^2+20x^4y^4+2x^2y^6-5y^8) - (4x^6y^2+20x^4y^4+14x^2y^6-4y^8)}{(x^2+y^2)^4}$
Oh, I guess I messed up here (assuming I did everything above correctly, anyways)

$= \frac{x^8+10x^6y^2-12x^2y^6-y^8}{(x^2+y^2)^4}$

But still, wouldn't the denominator become 0 at the origin?

6. Yes, the denominator will become 0 at the origin. This means Clairaut's theorem does not apply. However, I think you'll find that the mixed partial derivatives have the same form. So what is that telling you?

7. Hmm... Does this mean the results of the theorem can still occur when the theorem itself doesn't apply?

8. Yes. Think about the statement, "If A then B." This is the format of most theorems. B can be true whether or not A is or not. The situation you cannot have is when A is true and B is false. Any other truth combination of A and B can still hold.

9. Originally Posted by Ackbeet
Yes. Think about the statement, "If A then B." This is the format of most theorems. B can be true whether or not A is or not. The situation you cannot have is when A is true and B is false. Any other truth combination of A and B can still hold.
Ah, it's nice and clear now. Thanks a lot.

10. You're very welcome. Have a good one!