if f(x,y) = e^(-x)cosy + e^(-y) cosx
find (∂^2 f)/(∂x^2 )+ (∂^2 f)/(∂y^2 )- 2(∂^2 f)/∂x∂y
∂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
∂^f/∂x∂y = ∂^2f/∂y∂x = -e^(-x)siny + e^(-y)sinx
---> how we get this??
same way you get the other 2nd-order partials, but take derivative first on x, then on y.
∂^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...
To find (∂^2 f)/(∂x^2 )+ (∂^2 f)/(∂y^2 )- 2(∂^2 f)/∂x∂y, we need to calculate the second partial derivatives of f with respect to x and y.
Step 1: Calculate the first partial derivatives:
∂f/∂x = -e^(-x)cosy - e^(-y)sinx
∂f/∂y = -e^(-x)siny - e^(-y)cosx
Step 2: Calculate the second partial derivatives:
(∂^2 f)/(∂x^2) = e^(-x)cosy + e^(-y)cosx
(∂^2 f)/(∂y^2) = -e^(-x)siny - e^(-y)cosx
(∂^2 f)/(∂x∂y) = e^(-x)siny + e^(-y)sinx
Step 3: Substitute the results into the expression to find the final answer:
(∂^2 f)/(∂x^2 )+ (∂^2 f)/(∂y^2 )- 2(∂^2 f)/∂x∂y
= (e^(-x)cosy + e^(-y)cosx) + (-e^(-x)siny - e^(-y)cosx) - 2(e^(-x)siny + e^(-y)sinx)
= e^(-x)cosy - e^(-x)siny + e^(-y)cosx - e^(-y)siny - 2e^(-x)siny - 2e^(-y)sinx
This is the final expression for (∂^2 f)/(∂x^2 )+ (∂^2 f)/(∂y^2 )- 2(∂^2 f)/∂x∂y.