How do I find this partial derivative

Question

I have the following function u(x,y) defined as:

$$u(x,y) = \frac {xy(x^2-y^2)}{(x^2+y^2)}$$ when x and y are both non zero, and $u(0,0)=0$

I want to compute its partial derivative $u_{xy}$ at (0,0).

How do I do this?

Answer 1

You will need $\frac{\partial u}{\partial y}(x,y)$ and $\frac{\partial u}{\partial y}(0,0)$. The first one you can get using the quotient rule. And for the other, remember that by definition, we have: $$\frac{\partial u}{\partial y}(0,0) = \lim_{t \to 0} \frac{u(0, 0+t) - u(0,0)}{t}$$ With this in hands, we use the definition again: $$\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)(0,0) = \lim_{t \to 0} \frac{\frac{\partial u}{\partial y}(0+t,0) - \frac{\partial u}{\partial y}(0,0)}{t}.$$

Answer 2

$$\frac{\partial u}{\partial x}(0,0) = \lim_{t \to 0} \frac{u(0+t, 0) - u(0,0)}{t} = \lim_{t \to 0} \frac{0 - 0}{t}=0. $$

And we have

$$\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right)(0,0) = \lim_{t \to 0} \frac{\frac{\partial u}{\partial x}(0,0+t) - \frac{\partial u}{\partial x}(0,0)}{t}=\frac{-t}{t}=-1.$$