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Question
For the following functions find the fx, and fy and show that fxy = fyx
f(x, y) = `cos(x^2 - 3xy)`
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Solution
`(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)`
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