If 𝑧=𝑠𝑖𝑛⁡(3𝑥+2𝑦), determine 3∂2𝑧∂𝑦2-2∂2𝑧∂𝑥2

To solve this problem, we first need to find the second partial derivatives of z with respect to x and y.

The given function is z = sin(3x + 2y).

First, let's find the first partial derivatives:

∂z/∂x = cos(3x + 2y) * 3 ∂z/∂y = cos(3x + 2y) * 2

Now, let's find the second partial derivatives:

∂²z/∂x² = -sin(3x + 2y) * 3² = -9sin(3x + 2y) ∂²z/∂y² = -sin(3x + 2y) * 2² = -4sin(3x + 2y)

Finally, we substitute these into the given expression:

3∂²z/∂y² - 2∂²z/∂x² = 3(-4sin(3x + 2y)) - 2(-9sin(3x + 2y)) = -12sin(3x + 2y) + 18sin(3x + 2y) = 6sin(3x + 2y)