Question

If $u=\log (x^3+y^3+z^3-3xyz),$ then $(x,y,z)\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)$ is equal to:

• A. 0
• B. 1
• C. $u$
• D. 3
• E. $-1$

Answer 1

Answer: (D)

$\because$ $u=\log (x^3+y^3+z^3-3xyz)$ On differentiating w. r. t. $x,y,z$ respectively, we get $\frac{\partial u}{\partial x}=\frac{1}{x^3+y^3+z^3-3xyz}(3x^2-3yz)$ $\frac{\partial u}{\partial y}=\frac{1}{x^3+y^3+z^3-3xyz}(3y^2-3xz)$ and $\frac{\partial u}{\partial z}=\frac{1}{x^3+y^3+z^3-3xyz}(3z^2-3xy)$ $\therefore$ $(x+y+z)\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)$ $=\frac{3(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{(x^3+y^3+z^3-3xyz)}$ $=\frac{3(x^3+y^3+z^3-3xyz)}{(x^3+y^3+z^3-3xyz)}=3$