find the second-order partial derivatives of f(x,y)=x^3 + x^2y^2 + y^3 + x+y
and show that the mixed partial derivatives fxy and fyx are equal
Ah, the exciting world of second-order partial derivatives! Let's dive in and make things fun.
To find the second-order partial derivatives of f(x, y) = x^3 + x^2y^2 + y^3 + x + y, we'll have to take the partial derivatives twice. Buckle up, it might get a little bumpy.
First, let's find the first-order partial derivatives:
∂f/∂x = 3x^2 + 2xy^2 + 1
∂f/∂y = 2x^2y + 3y^2 + 1
Now we move on to the second-order partial derivatives. Let's find ∂^2f/∂x^2 first:
∂^2f/∂x^2 = ∂/∂x (3x^2 + 2xy^2 + 1)
= 6x + 2y^2
Now let's find ∂^2f/∂y^2:
∂^2f/∂y^2 = ∂/∂y (2x^2y + 3y^2 + 1)
= 2x^2 + 6y
Finally, let's find the mixed partial derivatives ∂^2f/∂x∂y and ∂^2f/∂y∂x:
∂^2f/∂x∂y = ∂/∂y (3x^2 + 2xy^2 + 1)
= 4xy
∂^2f/∂y∂x = ∂/∂x (2x^2y + 3y^2 + 1)
= 4xy
Oh, look at that! The mixed partial derivatives ∂^2f/∂x∂y and ∂^2f/∂y∂x are equal to 4xy. Sometimes, math can surprise you with its symmetry. It's like finding a matching pair of socks in your drawer - a rare but delightful occurrence.
So, in conclusion, the second-order partial derivatives of f(x, y) are:
- ∂^2f/∂x^2 = 6x + 2y^2
- ∂^2f/∂y^2 = 2x^2 + 6y
- ∂^2f/∂x∂y = ∂^2f/∂y∂x = 4xy
Hope that brought a smile to your face, or at least made you chuckle a little. If you have any more math questions or need a dose of humor, feel free to ask!
To find the second-order partial derivatives of the function f(x, y) = x^3 + x^2y^2 + y^3 + x + y, we need to start by finding the first-order partial derivatives, and then differentiate them again with respect to x and y.
1. Finding the first-order partial derivatives:
∂f/∂x = 3x^2 + 2xy^2 + 1
∂f/∂y = 2x^2y + 3y^2 + 1
2. Differentiating the first-order partial derivatives with respect to x:
∂^2f/∂x^2 = d/dx(∂f/∂x) = d/dx(3x^2 + 2xy^2 + 1) = 6x + 2y^2
3. Differentiating the first-order partial derivatives with respect to y:
∂^2f/∂y^2 = d/dy(∂f/∂y) = d/dy(2x^2y + 3y^2 + 1) = 2x^2 + 6y
4. Calculating the mixed partial derivatives:
∂^2f/∂x∂y = d/dx(∂f/∂y) = d/dx(2x^2y + 3y^2 + 1) = 4xy
∂^2f/∂y∂x = d/dy(∂f/∂x) = d/dy(3x^2 + 2xy^2 + 1) = 4xy
From step 4, we can observe that ∂^2f/∂x∂y and ∂^2f/∂y∂x are both equal to 4xy. Therefore, we have shown that the mixed partial derivatives fxy and fyx are equal for the function f(x, y) = x^3 + x^2y^2 + y^3 + x + y.