Final answer:
To find the first and second order partial derivatives of f(x, y) = -9sin(2x+y) - 9cos(x-y), differentiate with respect to x and y separately. The first order partial derivative with respect to x is -18 * cos(2x+y) + 9 * sin(x-y) and with respect to y is -9 * cos(2x+y) + 9 * sin(x-y). The second order partial derivatives can be found by differentiating the first order derivatives with respect to x and y.
Step-by-step explanation:
To find the first and second order partial derivatives of f(x, y) = -9sin(2x+y) - 9cos(x-y), we will differentiate with respect to x and y separately.
First, we find the first order partial derivative with respect to x (f_x):
f_x = -9 * (d/dx)sin(2x+y) - 9 * (d/dx)cos(x-y)
Using the chain rule and the derivative of sin(x) and cos(x), we can simplify this expression:
f_x = -18 * cos(2x+y) + 9 * sin(x-y)
Next, we find the first order partial derivative with respect to y (f_y):
f_y = -9 * (d/dy)sin(2x+y) + 9 * (d/dy)cos(x-y)
Using the chain rule and the derivative of sin(x) and cos(x), we can simplify this expression:
f_y = -9 * cos(2x+y) + 9 * sin(x-y)
Finally, we can find the second order partial derivatives by differentiating the first order derivatives with respect to x and y, separately.