If f(x,y) = {[(x^2)y+x(y^2)]*sin(x-y)}/(x^2 + y^2) ;(x,y)not=(0,0) & =0;(x,y)=(0,0). Find Fyx and Fxy at (0,0)

Question

what are Fxy and Fyx, partials?

Answer 1

Cleaning up first:

f(x,y)=[(yx^2 + xy^2)*sin(x-y)]/(x^2 + y^2) (right?)

To find Fyx and Fxy, you'll first need Fx and Fy.

Fx=[(x^2 + y^2)[(yx^2 + xy^2)*cos(x-y)+sin(x-y)*(2xy+y^2)] - [(yx^2 + xy^2)*sin(x-y)]*(2x)] /(x^2 + y^2)^2

Fy=[(x^2 + y^2)[(yx^2 + xy^2)*cos(x-y)(-1)+sin(x-y)*(x^2+2xy)] - [(yx^2 + xy^2)*sin(x-y)]*(2y)] /(x^2 + y^2)^2

Wow, those are nasty. I would highly recommend doing some cleanup algebra before taking the next partial.

Fx=[(yx^4+x^3y^2+x^2y^3+xy^4)cos(x-y) + (4x^2y+2xy^2-2x^3y-2x^2y^2)*sin(x-y)]/(x^2+y^2)^2

Fy=[(-x^4y-x^3y^2-x^2y^3-xy^4)*cos(x-y) + (x^2+2xy-2x^2y^2-2xy^3)*sin(x-y)]/(x^2+y^2)^2

Not that these are much cleaner, but there they are. Now you need to find (respectively) Fxy and Fyx.

Fxy=(x^2+y^2)^2*[(x^4+2x^3y+3x^2y^2+4xy^3)*cos(x-y) + sin(x-y)*(-1)*(yx^4+x^3y^2+x^2y^3+xy^4) + (4x^2+4xy-2x^3-4x^2y)*sin(x-y) + cos(x-y)*(4x^2y+2xy^2-2x^3y-2x^2y^2)] - [(yx^4+x^3y^2+x^2y^3+xy^4)*cos(x-y) + (4x^2y+2xy^2-2x^3y-2x^2y^2)*sin(x-y)]*(2*(x^2+y^2)*(2y)]/(x^2+y^2)^4

Can it be simplified? Yeah, but you're evaluating at (0,0), so why do it? You have polynomials in x & y in numerator and denominator, evaluated at (0,0) all go to 0. Thus, your answer is:
Fxy(0,0)=0/0

Goddamn that was a lot of work to show jack ****. I would assume, to show that there is an actual value, to choose a path over which to evaluate the partial (some path that will probably make the denominator non zero). If so, I would suggest the path x^2=y^2 only because it compacts the bottom term together as one, and it will most likely cancel out with the numerator making it possible to find a numeric value to this pseudo-limit.

I leave Fyx up to you.

Answer 2

Pi = delicous!

:)