Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct answer:

$u=\log (x^3+y^3+z^3-3xyz)$

Differentiating $u$ partially with $x,y,z$ respectively we get

$\Rightarrow$$\frac{\partial u}{\partial x}=\frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$$...\left(i\right)$

$\Rightarrow$$\frac{\partial u}{\partial y}=\frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$$...\left(ii\right)$

$\Rightarrow$$\frac{\partial u}{\partial z}=\frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$$...\left(iii\right)$

Adding $\left(i\right),\left(ii\right),\left(iii\right)$ we get

$\Rightarrow$$\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)=\frac{3x^2+3y^2+3z^2-3xy-3yz-3xz}{x^3+y^3+z^3-3xyz}$

$\Rightarrow$$\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)=\frac{3\left[\left(x^2+y^2+z^2\right)-\left(xy+yz+xz\right)\right]}{x^3+y^3+z^3-3xyz}$

On factorizing $x^3+y^3+z^3-3xyz$ we get

$x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left[\left(x^2+y^2+z^2\right)-\left(xy+yz+xz\right)\right]$$...\left(iv\right)$

From (iv) we can write

$\Rightarrow$ $\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)=\frac{3\left[\left(x^2+y^2+z^2\right)-\left(xy+yz+xz\right)\right]}{\left(x+y+z\right)\left[\left(x^2+y^2+z^2\right)-\left(xy+yz+xz\right)\right]}$

$\Rightarrow$ $\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)=\frac{3}{\left(x+y+z\right)}$

$\Rightarrow$$\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)\left(x+y+z\right)=3$

Hence, the value of $\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}\right)(x+y+z)$ is $3$, so option (D) is the correct answer.