Question

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

f(x, y) = $(3x)/(y + sinx)$

Answer 1

fx = $((y + sinx)[3] - 3x[0 + cos x])/(y + sin x)^2$

= $(3y + 3sinx - 3xcosx)/(y + sinx)^2$

fy = $((y + sin x)[0] - 3x[1 + 0])/(y + sinx)^2$

= $(- 3x)/(y + sinx)^2$

$(del^2"f")/(delxdely) = del/(delx)[(- 3x)/(y + sinx)^2]$

= $((y + sinx)^2 [- 3] - (- 3x)2(y + sinx)[0 + cosx])/(y + sinx)^4$

= $(- 3(y + sinx)^2 + 6x cosx(y + sinx))/(y + sinx)^4$ ........(1)

$(del^2"f")/(delydelx) = del/(dely) [(3y + 3sinx - 3x cosx)/(y + sinx)^2]$

= $((y + sinx)^2[3] - (3y + 3sinx)2(y + sinx)(0 + cosx))/(y + sinx)^4$

= $(-3(y + sinx)^2 + 6x cos x(y + sinx))/(y + sin x)^4$ ........(2)

From (1) and (2)

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