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Question: If \({\text{u = log(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3x...

If u = log(x3 + y3 + z3 - 3xyz){\text{u = log(}}{{\text{x}}^3}{\text{ + }}{{\text{y}}^3}{\text{ + }}{{\text{z}}^3}{\text{ - 3xyz)}} and (∂∂x + ∂∂y + ∂∂z)2u = −k(x + y + z)2{\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

Explanation

Solution

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 u = log(x3 + y3 + z3 - 3xyz){\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
∂u∂x = 3x2 - 3yz(x3 + y3 + z3 - 3xyz)\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
∂u∂y = 3y2 - 3xz(x3 + y3 + z3 - 3xyz)\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)}}}}
∂u∂z = 3z2 - 3xy(x3 + y3 + z3 - 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
∂u∂x + ∂u∂y + ∂u∂z = 3x2 + 3y2 + 3z2 - 3xy - 3yz - 3xz(x3 + y3 + z3 - 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{ + 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)}}}}
∂u∂x + ∂u∂y + ∂u∂z = 3(x2 + y2 + z2) - 3(xy + yz + xz)(x3 + y3 + z3 - 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)}}}}
∂u∂x + ∂u∂y + ∂u∂z = 3(x2 + y2 + z2 - xy - yz - xz)(x3 + y3 + z3 - 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 x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz){{\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, ∂u∂x + ∂u∂y + ∂u∂z = 3(x2 + y2 + z2 - xy - yz - xz)(x + y + z)(x2 + y2 + z2 - xy - yz - xz)\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 ∂u∂x + ∂u∂y + ∂u∂z = 3(x + y + z)\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, (∂∂x + ∂∂y + ∂∂z)(∂u∂x + ∂u∂y + ∂u∂z) = (∂∂x + ∂∂y + ∂∂z)2u\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}}
∂∂x(∂u∂x + ∂u∂y + ∂u∂z) = ∂∂x(3(x + y + z))\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) = −3(x + y + z)2\dfrac{{ - 3}}{{{{({\text{x + y + z)}}}^2}}}
Similarly, we get
∂∂y(∂u∂x + ∂u∂y + ∂u∂z) = −3(x + y + z)2\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 ∂∂z(∂u∂x + ∂u∂y + ∂u∂z) = −3(x + y + z)2\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 (∂∂u + ∂∂y + ∂∂z)2u = −3(x + y + z)2 + −3(x + y + z)2 + −3(x + y + z)2{\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, (∂∂u + ∂∂y + ∂∂z)2u = −9(x + y + z)2{\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 (∂∂x + ∂∂y + ∂∂z)2u = −k(x + y + z)2{\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 .