Question

Let the function f:RR²→RR be defined as f(x,y)={xy(x²-y²)/x²+y², (x,y)≠(0,0) {0, (x,y)=(0,0) Show that (i)f_x(0,y)=y,for all y (ii)f_x(x,0)=x,for all x. Hence verify that f_xy(0,0)≠f_yx(0,0).?

Answer 1

`f_x(0,y) = -y,qquad forall y`
`f_y(x,0) = x, qquad forall x`
`f_{xy} (0,0)= -1`
`f_{yx}(0,0) = 1`

Explanation:

To calculate the derivative at all points other than `(0,0)` is straightforward.

If `y ne 0, forall x`, we have

`f_x = {(x^2+y^2){y(x^2-y^2)-2x^2y}-xy(x^2-y^2)2x}/{(x^2+y^2)^2}`

Since we need the value of the derivative for `x = 0`, it simplifies considerably :
`f_x(0,y) = {-y^5}/{y^4}=-y`
Note that this works only when `y ne 0` For `(x,y) = (0,0)`, we need to calculate the derivative from first principles :

`f_x(0,0) = lim_{h to 0} {f(h,0)-f(0,0)}/{h} = lim_{h to 0}{0-0}/h=0`

Combining the two results, we get

`f_x(0,y) = -y,qquad forall y`

In a similar manner, we can show that

`f_y(x,0) = x, qquad forall x`

Note that `f_{xy}(0,0) = partial/{partial y} f_x(0,y)|_(0,0)` so we only need to know `f_y(0,y)` to determine this - the behavior of the function `f_x` at other values of `x` is not needed. It is easy to see that
`f_{xy} (0,0)= -1`
and similarly
`f_{yx}(0,0) = 1`