Question

If $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$, then find the value of $\left( {\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} + \dfrac{{\partial u}}{{\partial z}}} \right)\left( {x + y + z} \right)$.
(A) $0$
(B) $1$
(C) $2$
(D) $3$

Answer 1

Hint: In the given problem, we are required to differentiate the given expression partially with three variables and then find the value of the expression given to us. Since, $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$ is a composite function, so we will have to apply chain rule of differentiation in the process of differentiating $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$ . So, differentiation of $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$ with respect to any of three variables will be done layer by layer using the chain rule of differentiation. Also derivatives of $\ln \left( x \right)$ with respect to $x$ must be remembered to solve the given problem.

Complete answer: So, we have, $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$. So, we need to find the partial derivatives of the function. So, we partially differentiate the function with respect to the variables x, y and z separately and then substitute their values in the function to evaluate the expression $\left( {\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} + \dfrac{{\partial u}}{{\partial z}}} \right)\left( {x + y + z} \right)$.

Partial differentiation with respect to a variable is done considering all the other variables as constant.
So, differentiating the function $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$ with respect to x, we consider y and z to be constants.
So, we get, $\dfrac{{\partial u}}{{\partial x}} = \dfrac{\partial }{{\partial x}}\left[ {\log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)} \right]$
Now, using the chain rule of differentiation, we get,
$ \Rightarrow \dfrac{{\partial u}}{{\partial x}} = \dfrac{1}{{\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}} \times \left( {3{x^2} - 3yz} \right)$
We know that the derivative of logarithmic function is $\left( {\dfrac{1}{x}} \right)$.
Similarly, differentiating the function $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$ with respect to y, we consider x and z to be constants.
So, we get, $\dfrac{{\partial u}}{{\partial y}} = \dfrac{\partial }{{\partial y}}\left[ {\log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)} \right]$
Now, using the chain rule of differentiation, we get,
$ \Rightarrow \dfrac{{\partial u}}{{\partial y}} = \dfrac{1}{{\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}} \times \left( {3{y^2} - 3xz} \right)$
Similarly, differentiating the function $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$ with respect to z, we consider x and y to be constants.
So, we get, $\dfrac{{\partial u}}{{\partial z}} = \dfrac{\partial }{{\partial z}}\left[ {\log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)} \right]$
Now, using the chain rule of differentiation, we get,
$ \Rightarrow \dfrac{{\partial u}}{{\partial z}} = \dfrac{1}{{\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}} \times \left( {3{z^2} - 3xy} \right)$
Now, substituting the values of $\dfrac{{\partial u}}{{\partial x}}$, $\dfrac{{\partial u}}{{\partial y}}$ and $\dfrac{{\partial u}}{{\partial z}}$ in the expression $\left( {\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} + \dfrac{{\partial u}}{{\partial z}}} \right)\left( {x + y + z} \right)$, we get,
$ \Rightarrow \left( {\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} + \dfrac{{\partial u}}{{\partial z}}} \right)\left( {x + y + z} \right)$
$ \Rightarrow \left( {\dfrac{{\left( {3{x^2} - 3yz} \right)}}{{\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}} + \dfrac{{\left( {3{y^2} - 3xz} \right)}}{{\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}} + \dfrac{{\left( {3{z^2} - 3xy} \right)}}{{\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}}} \right)\left( {x + y + z} \right)$
Now, adding up all the terms in numerator, we get,
$ \Rightarrow \left( {\dfrac{{3{x^2} - 3yz + 3{y^2} - 3xz + 3{z^2} - 3xy}}{{\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}}} \right)\left( {x + y + z} \right)$
Now, taking $3$ common from numerator and using the algebraic identity $\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right) = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - yz - zx - xy} \right)$ in the numerator, we get,
$ \Rightarrow 3\left( {\dfrac{{{x^2} - yz + {y^2} - xz + {z^2} - xy}}{{\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}}} \right)\left( {x + y + z} \right)$
$ \Rightarrow 3\left( {\dfrac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{{x^3} + {y^3} + {z^3} - 3xyz}}} \right)$
Now, cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow 3$
So, the value of $\left( {\dfrac{{\partial u}}{{\partial x}} + \dfrac{{\partial u}}{{\partial y}} + \dfrac{{\partial u}}{{\partial z}}} \right)\left( {x + y + z} \right)$ is $3$ for $u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$.
Hence, option (D) is the correct answer.

Note:
The given problem involves the knowledge of partial derivatives with respect to a variable. The derivatives of basic functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. We must know the algebraic identity for the simplification of the expression involving partial derivatives.