# Partial Derivative which when evaluated gives $0/0$

My function is $f(x,y)=xy\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ EDIT defined $f(0,0)=0$ (How does that impact the evaluation of partial derivatives?)

I need to evaluate the function at $f_{xy}(0,0)$ and similarly for $f_{yx}(0,0)$ My problem is that upon evaluating the function at (0,0), I get cases of $0/0$ How can I get around it? Should I try rearranging terms with algebra?

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Your function is not defined at (0,0). You might try evaluating what happens in the limit as you approach the origin along any line. – Arkamis Nov 28 '12 at 16:24
Sorry, I should have mentioned that the function is given as defined at $f(0,0)=0$ But I didn't think this had relevance for the partial derivatives. – Edgar Aroutiounian Nov 28 '12 at 16:47
...evaluate teh function at $f_{xy}(0,0)$, I think is meaningless. Maybe you want to verify if $f(x,y)$ has $f_{xy}$ and $f_{yx}$ at the origin? – Babak S. Nov 28 '12 at 16:51

You'll need to use the definition of derivative to find $f'_x$ and $f'_y$ at $(0,0)$; for $(x,y)\neq (0,0)$ you just use differentiate the expression for $f$ as usual. Then you have formulas $f'_x(x,y)$ equal to something at $(0,0)$ and something else elsewhere, and similarly for $f'_y(x,y)$. Then you again use the definition of derivative on those functions to get $f''_{xy}$ and $f''_{yx}$ at $(0,0)$.
The expressions you get by differentiating $xy(x^2-y^2)/(x^2+y^2)$ only say what the derivatives are for $(x,y) \neq (0,0)$; it's completely meaningless to try to substitute $x=y=0$ into those formulas, if that's what you're doing when you're getting 0/0. To find $f'_x(0,0)$, you must do an entirely different computation, namely calculate $\lim_{h \to 0} \frac{1}{h}(f(h,0)-f(0,0))$. – Hans Lundmark Nov 28 '12 at 20:26
Let $y=ax$ and take the limit as $x \rightarrow 0$.
Edit: If the result of taking the above limit is not a constant but a function of $a$, the limit does not exist. You can, however, find the value of $f_{xy}(0,0)$ or $f_{yx}(0,0)$ from a particular direction by picking $a$ such that $y=ax$ defines the line along which you want to approch $(0,0)$.