Why of $f_1, f_2$ exists everywhere for $\frac{x*y}{x^2+y^2}$?

Question

Let $f(x,y):R^2\rightarrow R$ to be the function such that $f(x,y)=\frac{x*y}{x^2+y^2}$?

Why of $f_1, f_2$ exists everywhere for $\frac{x*y}{x^2+y^2}$?

For example $f_1=f_x=\frac {\partial f(x,y)}{x}=\frac{y(x^2+y^2)-xy(2*x)}{(x^2+y^2)^2}$. When I consider the point where $(x,y)=(0,0)$, I didn't see why $f_1$ still exists.

Answer 1

In this case you know that If you think about the definition of the derivative it's a bit clearer. Firstly the function is not well defined. I assume it should say: \begin{align*} f(x,y) = \begin{cases} 0 & (x,y)=(0,0)\\ \frac{xy}{x^2+y^2} & \mathrm{otherwise}\\ \end{cases} \end{align*} Now by applying the definition of partial differentiation, \begin{align*} f_1(0,0) = \frac{\partial f(x, y)}{\partial x}\Big|_{(0,0)} = \lim_{h\rightarrow0}\frac{f(0+h, 0) - f(0, 0)}{h} = \lim_{h\rightarrow0}\frac{\frac{0h}{h^2+0}-0}{h} = \lim_{h\rightarrow0}\frac{0}{h^3} = 0 \end{align*}

Similarly you can show that $f_2$ exists.

Answer 2

You have to answer a question like this by returning to the original limit definition of the partial derivative. In other words, you get this result by returning to the first principle definition.