Question

if f(x,y) = e^(-x)cosy + e^(-y) cosx
find (∂^2 f)/(∂x^2 )+ (∂^2 f)/(∂y^2 )- 2(∂^2 f)/∂x∂y

Answer 1

∂f/∂x = -e^(-x)cosy - e^(-y)sinx
∂^2f/∂x^2 = e^(-x)cosy - e^(-y)cosx

∂f/∂y = -e^(-x)siny - e^(-y)cosx
∂^2f/∂y^2 = -e^(-x)cosy + e^(-y)cosx

∂^f/∂x∂y = ∂^2f/∂y∂x = -e^(-x)siny + e^(-y)sinx

so, we wind up with

e^(-x)cosy - e^(-y)cosx
+(-e^(-x)cosy + e^(-y)cosx)
-2(-e^(-x)siny + e^(-y)sinx)
=
e^(-x)[cosy-cosy+2siny] + e^(-y)[-cosx+cosx-2sinx]
= 2e^(-x)siny - 2e^(-y)sinx

As always, check my algebra

Answer 2

∂^f/∂x∂y = ∂^2f/∂y∂x = -e^(-x)siny + e^(-y)sinx


---> how we get this??

Answer 3

same way you get the other 2nd-order partials, but take derivative first on x, then on y.

Answer 4

∂^f/∂x∂y = ∂^2f/∂y∂x = -e^(-x)siny + e^(-y)sinx

if i'm not mistaken...exist -e^(-x)siny is positive ( e^(-x)siny ) not negative... coz y difftation -siny and difftation -x is -1.. so (-)(-) is positive...