Question

If ${\text{u = log(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}}$ and ${\left( {\dfrac{\partial }{{\partial {\text{x}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{y}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{z}}}}} \right)^2}{\text{u = }}\dfrac{{ - {\text{k}}}}{{{{({\text{x + y + z)}}}^2}}}$, then k =?
A. 6
B. 3
C. 9
D. 5

Answer 1

Hint: To solve this question, we will use partial differentiation to differentiate the given function. In Partial differentiation if we are differentiating a function which is a product of x and y, so to differentiate with respect to x, we will keep y as a constant and differentiate the function with respect to x. In this question, we will partially differentiate u two times to find the value of k.

Complete step-by-step answer:

Now, we will use partial differentiation. It is denoted by $\partial $. We are given ${\text{u = log(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}}$. So, partially differentiating u with respect to x, we get
$\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ = }}\dfrac{{3{{\text{x}}^2}{\text{ - 3yz}}}}{{{\text{(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}}}}$
Similarly, partially differentiating u with respect to y and z, we get
$\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ = }}\dfrac{{3{{\text{y}}^2}{\text{ - 3xz}}}}{{{\text{(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}}}}$
$\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}{\text{ = }}\dfrac{{3{{\text{z}}^2}{\text{ - 3xy}}}}{{{\text{(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}}}}$
On adding all the three partial derivatives, we get
$\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}{\text{ = }}\dfrac{{3{{\text{x}}^2}{\text{ + 3}}{{\text{y}}^2}{\text{ + 3}}{{\text{z}}^2}{\text{ - 3xy - 3yz - 3xz}}}}{{{\text{(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}}}}$
$\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}{\text{ = }}\dfrac{{3({{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ + }}{{\text{z}}^2}{\text{) - 3(xy + yz + xz)}}}}{{{\text{(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}}}}$
$\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}{\text{ = }}\dfrac{{3({{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ + }}{{\text{z}}^2}{\text{ - xy - yz - xz)}}}}{{{\text{(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}}}}$
Now, we know that ${{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz = (x + y + z)(}}{{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ + }}{{\text{z}}^2}{\text{ - xy - yz - xz)}}$
Therefore, $\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}{\text{ = }}\dfrac{{3({{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ + }}{{\text{z}}^2}{\text{ - xy - yz - xz)}}}}{{({\text{x + y + z)}}({{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ + }}{{\text{z}}^2}{\text{ - xy - yz - xz)}}}}$ $\Rightarrow$ $\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}{\text{ = }}\dfrac{3}{{({\text{x + y + z)}}}}$
Now, $\left( {\dfrac{\partial }{{\partial {\text{x}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{y}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{z}}}}} \right)\left( {\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}} \right){\text{ = }}{\left( {\dfrac{\partial }{{\partial {\text{x}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{y}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{z}}}}} \right)^2}{\text{u}}$
$\dfrac{\partial }{{\partial {\text{x}}}}\left( {\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}} \right){\text{ = }}\dfrac{\partial }{{\partial {\text{x}}}}\left( {\dfrac{3}{{({\text{x + y + z)}}}}} \right)$ = $\dfrac{{ - 3}}{{{{({\text{x + y + z)}}}^2}}}$
Similarly, we get
$\dfrac{\partial }{{\partial {\text{y}}}}\left( {\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}} \right){\text{ = }}\dfrac{{ - 3}}{{{{({\text{x + y + z)}}}^2}}}$ and $\dfrac{\partial }{{\partial {\text{z}}}}\left( {\dfrac{{\partial {\text{u}}}}{{\partial {\text{x}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{y}}}}{\text{ + }}\dfrac{{\partial {\text{u}}}}{{\partial {\text{z}}}}} \right){\text{ = }}\dfrac{{ - 3}}{{{{({\text{x + y + z)}}}^2}}}$
So, we get ${\left( {\dfrac{\partial }{{\partial {\text{u}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{y}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{z}}}}} \right)^2}{\text{u = }}\dfrac{{ - 3}}{{{{({\text{x + y + z)}}}^2}}}{\text{ + }}\dfrac{{ - 3}}{{{{({\text{x + y + z)}}}^2}}}{\text{ + }}\dfrac{{ - 3}}{{{{({\text{x + y + z)}}}^2}}}$
Therefore, ${\left( {\dfrac{\partial }{{\partial {\text{u}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{y}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{z}}}}} \right)^2}{\text{u = }}\dfrac{{ - 9}}{{{{({\text{x + y + z)}}}^2}}}$
Comparing it with ${\left( {\dfrac{\partial }{{\partial {\text{x}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{y}}}}{\text{ + }}\dfrac{\partial }{{\partial {\text{z}}}}} \right)^2}{\text{u = }}\dfrac{{ - {\text{k}}}}{{{{({\text{x + y + z)}}}^2}}}$, we get
k = 9
So, Option (C) is the correct answer.

Note: When we come up with such types of questions, we have to use partial differentiation to solve the problem. The rules to do partial differentiation of a function is the same as that of normal differentiation. In partial differentiation we keep one variable as constant and we differentiate the other variable. Most of the students make a mistake while solving such types of questions. They do normal differentiation instead of partial differentiation because they get confused between the sign of integration. Normal integration is represented by d and partial differentiation is represented by $\partial $.