Hi! Given the function (2xy(x^2-y^2))/(x^2+y^2). The function is defined everywh
ID: 2842072 • Letter: H
Question
Hi!
Given the function (2xy(x^2-y^2))/(x^2+y^2).
The function is defined everywhere, except at (x,y)=(0,0). Here we put F(0,0)=0.
1) Calculate F_1(x,y), F_2 (x,y), F_12 and F_21 at points, where (x,y) is not equal to zero.
2) Calculate these dericatives at (0,0).
3) Observe that F_12(0,0)=2 and F_21(x,y) = -2. Does this contradict the Theorem: "Suppose that two miced n'th-orden partial derivates of a function f involve the samme differentiations but in different orders. If those partials are continuous at a point P, and if f and all partials of j of order less than n are continuous in a neighbourhood of P, then the two mixed partials are equal at the point P"?
4) Explain why!
Explanation / Answer
(2)
The derivatives at (0,0) may not be calculated because F is discontinous at (0,0).
Denominator in the terms will be 0 at (0,0). If it cancels somehow, then also the values F_12 may not be equal to F_21.
The answer for 2nd is simply not defined.
3rd part is the thing what I am saying. Due to discontinuity values are simply not defined at (0,0).
If you want other answers or any clarification, you can ask any time.