Question

Find all the first and second order partial derivatives of f(x, y) = 2sin(2x + y) + 6cos(x - y). A. ∂f/∂x = ∂f/∂y = ∂f/∂x = fx =

Answer 1

The first-order partial derivatives of the given function are ∂f/∂x = 4cos(2x + y) - 6sin(x - y) and ∂f/∂y = 2cos(2x + y) + 6sin(x - y). The second order partial derivatives are ∂²f/∂x² = -8sin(2x + y) - 6cos(x - y), ∂²f/∂y² = -4sin(2x + y) + 6cos(x - y), and ∂²f/∂x∂y = 4cos(2x + y) - 6sin(x - y).

Step-by-step explanation:

To find the first and second-order partial derivatives of the function f(x, y) = 2sin(2x + y) + 6cos(x - y), we differentiate the function concerning x and y separately.

First-order partial derivatives:

• ∂f/∂x = 4cos(2x + y) - 6sin(x - y)
• ∂f/∂y = 2cos(2x + y) + 6sin(x - y)

Second-order partial derivatives:

• ∂²f/∂x² = -8sin(2x + y) - 6cos(x - y)
• ∂²f/∂y² = -4sin(2x + y) + 6cos(x - y)
• ∂²f/∂x∂y = 4cos(2x + y) - 6sin(x - y)