Question

For the following functions find the f_x, and f_y and show that f_xy = f_yx 

f(x, y) = `cos(x^2 - 3xy)`

Answer 1

`(del"f")/(delx) = - sin(x^2 - 3xy) [2x - 3y]`

= `(3y - 2x) sin(x^2 - 3xy)`

`(del"f")/(dely) = - sin(x^2 - 3xy)[0 - 3x]`

= `3x sin(x^2 - 3xy)`

`(del^2"f")/(delxdely) = del/(delx)[(del"f")/(dely)]`

=`del/(delx) [3x sin(x^2 - 3xy)]`

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

= `3x(2x - 3y) cos(x^2 - 3xy) + 3 sin(x^2 - 3xy)` ........(1)

`(del^2"f")/(delydelx) = del/(dely) [(del"f")/(delx)]`

= `el/(dely) [(3y - 2x) sin(x^2 - 3xy)]`

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

 `3x (2x - 3y) cos(x^2 - 3xy) + 3sin (x^2 - 3xy)`  ........(2)

From (1) and (2)

⇒ `(del^2"f")/(delxdely) = (del^2"f")/(delydelx)`