Question

For the following functions find the fx, and fyand show that fxy= fyx

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)

$(del^2"f")/(delxdely) = (del^2"f")/(delydelx)$