Final answer:
The second-order partial derivatives of the function r(x,y) = ln(3x + y) are found to be r₂ₓₜ₂(x,y) = -9/(3x + y)², r₂ₓₜₚ(x,y) = -3/(3x + y)² and rₓₚₓₚ(x,y) = -1/(3x + y)². Clairaut's Theorem confirms the equality of the mixed partial derivatives.
Step-by-step explanation:
Finding Second-Order Partial Derivatives
The function given is r(x,y) = ln(3x + y). To find the second-order partial derivatives, we first find the first-order derivatives with respect to x and y, then differentiate each of these again with respect to x and y, respectively. The first order partial derivative with respect to x is given by:
rx(x,y) = ∂r/ ∂x = 1/(3x + y) * 3 = 3/(3x + y)
The first order partial derivative with respect to y is given by:
ry(x,y) = ∂r/ ∂y = 1/(3x + y)
Now, we differentiate these first-order derivatives again to find the second-order partial derivatives:
rxx(x,y) = ∂²r/ ∂x² = -9/(3x + y)²
rxy(x,y) = ∂²r/ ∂x∂y = -3/(3x + y)²
ryy(x,y) = ∂²r/ ∂y² = -1/(3x + y)²
Note that the mixed partial derivative rxy is equal to the mixed partial derivative ryx, as predicted by Clairaut's Theorem which states that the mixed derivatives will be equal if they are continuous.