Question

Find the correct option
If $u = \log ({x^3} + {y^3} + {z^3} - 3xyz)$ , then $\left(\dfrac{{du}}{{dx}} + \dfrac{{du}}{{dy}} + \dfrac{{du}}{{dz}}\right)(x + y + z)$ is equal to
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: Using relevant differentiation rules of logarithm function and algebraic function in the form of $x^n$, each derivative given in the expression is calculated separately one by one and putting those calculated values, the expression is simplified using suitable algebraic formulae of factorization to find the final value of the expression and to choose the correct option.

Formula Used:
The following differentiation and algebraic formulas have been used to simplify the given expression:
1. $\dfrac{{d(\log v)}}{{dx}} = \dfrac{1}{v} \times \dfrac{{dv}}{{dx}}$ , where, $v$ is a function of $x$.
2. $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$
3. ${x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - yz - zx)$

Complete step by step solution:
We have been given that $u = \log ({x^3} + {y^3} + {z^3} - 3xyz)$ .
We have to find out the value of the expression $\left( {\dfrac{{du}}{{dx}} + \dfrac{{du}}{{dy}} + \dfrac{{du}}{{dz}}} \right)(x + y + z)$ .
The above expression comprises three derivatives and one algebraic expression.
We will find out the value of each derivative one by one.
First, to find $\dfrac{{du}}{{dx}}$ , we will consider $y,z$ as constants and $u$ as a function of $x$ .
So, $\dfrac{{du}}{{dx}} = \dfrac{{d[\log ({x^3} + {y^3} + {z^3} - 3xyz)}}{{dx}}$
Taking ${x^3} + {y^3} + {z^3} - 3xyz$ equal to $v$, we have
$\dfrac{{du}}{{dx}} = \dfrac{{d(\log v)}}{{dx}} = \dfrac{1}{v} \times \dfrac{{dv}}{{dx}}$
Substituting the value of $v$ , we get
$\dfrac{{du}}{{dx}} = \dfrac{1}{{({x^3} + {y^3} + {z^3} - 3xyz)}} \times \dfrac{{d({x^3} + {y^3} + {z^3} - 3xyz)}}{{dx}}$
  $ = \dfrac{1}{{({x^3} + {y^3} + {z^3} - 3xyz)}} \times (3{x^2} - 3yz)$ [Applying the formula $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$ , where $n = 3$ ]
  $ = \dfrac{{3({x^2} - yz)}}{{({x^3} + {y^3} + {z^3} - 3xyz)}}$
Similarly, to find $\dfrac{{du}}{{dy}}$ , we will consider $z,x$ as constants and $u$ as a function of $y$ .
So, $\dfrac{{du}}{{dy}} = \dfrac{{d[\log ({x^3} + {y^3} + {z^3} - 3xyz)}}{{dy}}$
Taking ${x^3} + {y^3} + {z^3} - 3xyz$ equal to $v$, we have
$\dfrac{{du}}{{dy}} = \dfrac{{d(\log v)}}{{dy}} = \dfrac{1}{v} \times \dfrac{{dv}}{{dy}}$
Substituting the value of $v$ , we get
$\dfrac{{du}}{{dy}} = \dfrac{1}{{({x^3} + {y^3} + {z^3} - 3xyz)}} \times \dfrac{{d({x^3} + {y^3} + {z^3} - 3xyz)}}{{dy}}$
  $ = \dfrac{1}{{({x^3} + {y^3} + {z^3} - 3xyz)}} \times (3{y^2} - 3zx)$ [Applying the formula $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$ , where $n = 3$ ]
  $ = \dfrac{{3({y^2} - zx)}}{{({x^3} + {y^3} + {z^3} - 3xyz)}}$
And to find $\dfrac{{du}}{{dz}}$ , we will consider $x,y$ as constants and $u$ as a function of $z$ .
So, $\dfrac{{du}}{{dz}} = \dfrac{{d[\log ({x^3} + {y^3} + {z^3} - 3xyz)}}{{dz}}$
Taking ${x^3} + {y^3} + {z^3} - 3xyz$ equal to $v$, we have
$\dfrac{{du}}{{dz}} = \dfrac{{d(\log v)}}{{dz}} = \dfrac{1}{v} \times \dfrac{{dv}}{{dz}}$
Substituting the value of $v$ , we get
$\dfrac{{du}}{{dz}} = \dfrac{1}{{({x^3} + {y^3} + {z^3} - 3xyz)}} \times \dfrac{{d({x^3} + {y^3} + {z^3} - 3xyz)}}{{dz}}$
  $ = \dfrac{1}{{({x^3} + {y^3} + {z^3} - 3xyz)}} \times (3{z^2} - 3xy)$ [Applying the formula $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$ , where $n = 3$ ]
  $ = \dfrac{{3({z^2} - xy)}}{{({x^3} + {y^3} + {z^3} - 3xyz)}}$
Now, substituting the above calculated values of the derivatives in the expression, we have
$\left( {\dfrac{{du}}{{dx}} + \dfrac{{du}}{{dy}} + \dfrac{{du}}{{dz}}} \right)(x + y + z) = \left[ {\dfrac{{3({x^2} - yz)}}{{({x^3} + {y^3} + {z^3} - 3xyz)}} + \dfrac{{3({y^2} - zx)}}{{({x^3} + {y^3} + {z^3} - 3xyz)}} + \dfrac{{3({z^2} - xy)}}{{({x^3} + {y^3} + {z^3} - 3xyz)}}} \right] \times (x + y + z)$$ = \dfrac{{3({x^2} + {y^2} + {z^2} - xy - yz - zx)}}{{({x^3} + {y^3} + {z^3} - 3xyz)}} \times (x + y + z)$
$ = \dfrac{{3({x^3} + {y^3} + {z^3} - 3xyz)}}{{({x^3} + {y^3} + {z^3} - 3xyz)}}$ [Since ${x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - yz - zx)$]
$= 3$
So, the expression $\left( {\dfrac{{du}}{{dx}} + \dfrac{{du}}{{dy}} + \dfrac{{du}}{{dz}}} \right)(x + y + z)$ is equal to 3.

Option ‘D’ is correct

Note: The formula $\dfrac{{d(\log v)}}{{dx}} = \dfrac{1}{v} \times \dfrac{{dv}}{{dx}}$ is applied to find the derivative of the logarithm function in case of base $e$ , but, when, the base is different say $a$, then the formula applied will be $\dfrac{{d(\log v)}}{{dx}} = \dfrac{1}{v} \times \dfrac{1}{{\log a}} \times \dfrac{{dv}}{{dx}}$.