Final answer:
The calculation for the first function with given conditions does not match the expected result, potentially due to a typo. The second function's partial derivatives sum up to the square of the sum of variables. For the third function, the partial derivatives are found using the chain rule applied to the product of variables as a single variable.
Step-by-step explanation:
Mathematical Demonstration
1. For the function V = x² + y² + z², we want to show that xVx + yVy + zVz = 2πV. Assuming Vx, Vy, and Vz refer to the partial derivatives of V with respect to x, y, and z respectively, we have Vx = 2x, Vy = 2y, and Vz = 2z. Thus, xVx + yVy + zVz = 2x² + 2y² + 2z². Since this is twice the original expression for V, we find that xVx + yVy + zVz = 2( x² + y² + z²) = 2V, which does not equal 2πV. It seems there may be a typo in the given equation.
2. For the function u = xy² + yz² + zx², to show that u_x + u_y + u_z = (x + y + z)², we calculate the partial derivatives: u_x = y² + z², u_y = 2xy + z², and u_z = 2yz + x². Adding these, we get u_x + u_y + u_z = x² + y² + z² + 2xy + 2yz + 2zx, which simplifies to (x + y + z)².
3. For a function u of the form u = f(xyz), the partial derivatives u_x, u_y, u_z are found by applying the chain rule considering xyz as the variable. This yields u_x = f'(xyz) • yz, u_y = f'(xyz) • xz, and u_z = f'(xyz) • xy.