I know how to compute partial derivatives of functions with more than one variable. But how can i assert that the partial derivatives of a given function exist at a point without computing it? Consider the following functions for example:
$f(x,y)=xy(x^2-y^2)/(x^2+y^2)$ and $f(0,0)=0$
How do I show that the first partial derivative of $f(x,y)$ exists at $0$?
Thanks for you help in advance!
Hint: There are two first partials, but by (anti)symmetry you only need to show that one exists. So write it out as it is defined: show that $$\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h} $$ exists. Note that before you can do this, you have to define what $$ f(0,0)$$ is--your definition as stated doesn't include that point.
Addendum: Since you added the definition that $f(0)=0$, the problem becomes easier. The limit in question is now $$\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h} =\lim_{h\to 0}\frac{f(h,0)}{h}.$$ But note that for any nonzero $h$, we have $f(h,0)=(h)(0)(h^2-0^2)/(h^2+0^2)=0$, so this is just $$\lim_{h\to 0}\frac{0}{h}=\lim_{h\to 0}0 =0.$$
In other words, we showed $$\frac{\partial f}{\partial x}(0,0)=0.$$ By antisymmetry, we also then have that $$\frac{\partial f}{\partial y}(0,0)=0$$ as well.