Solveeit Logo
Login

Question

Question: Find the value of \(\left( {x + y + z} \right)\left( {{u_x} + {u_y} + {u_z}} \right)\) when \(u = \l...

Find the value of (x+y+z)(ux+uy+uz)\left( {x + y + z} \right)\left( {{u_x} + {u_y} + {u_z}} \right) when u=log⁡(x3+y3+z3−3xyz)u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)
(a) 0\left( a \right){\text{ 0}}
(b) x - y + z\left( b \right){\text{ x - y + z}}
(c) 2\left( c \right){\text{ 2}}
(d) 3\left( d \right){\text{ 3}}

Explanation

Solution

First of all we will find the derivative of u=log⁡(x3+y3+z3−3xyz)u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right) with respect to x,y,zx, y, z for ux,uy,uz{u_x}, {u_y}, {u_z} respectively. After that, we will add all three equations, and on solving we will get the values.

Formula used:
As we know
zz x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−xz){x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)
Differentiation formula,
dlog⁡xdx=1x\dfrac{{d\log x}}{{dx}} = \dfrac{1}{x}
dxndx=nxn−1\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}

Complete step by step solution:
As we have the equation given as u=log⁡(x3+y3+z3−3xyz)u = \log \left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)
Now on differentiating with respect to xx, we get
⇒ux=dudx=3x2−3yzx3+y3+z3−3xyz\Rightarrow {u_x} = \dfrac{{du}}{{dx}} = \dfrac{{3{x^2} - 3yz}}{{{x^3} + {y^3} + {z^3} - 3xyz}}, we will name it equation 11
Similarly, on differentiating with respect to yy, we get
⇒uy=dudy=3y2−3xzx3+y3+z3−3xyz\Rightarrow {u_y} = \dfrac{{du}}{{dy}} = \dfrac{{3{y^2} - 3xz}}{{{x^3} + {y^3} + {z^3} - 3xyz}}, we will name it equation 22
Similarly, on differentiating with respect to zz, we get
⇒uz=dudz=3z2−3xyx3+y3+z3−3xyz\Rightarrow {u_z} = \dfrac{{du}}{{dz}} = \dfrac{{3{z^2} - 3xy}}{{{x^3} + {y^3} + {z^3} - 3xyz}}, we will name it equation 33
Now on adding all these three equations, we get
⇒dudx+dudy+dudz=3(x2+y2+z2−xy−yz−zx)x3+y3+z3−3xyz\Rightarrow \dfrac{{du}}{{dx}} + \dfrac{{du}}{{dy}} + \dfrac{{du}}{{dz}} = \dfrac{{3\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)}}{{{x^3} + {y^3} + {z^3} - 3xyz}}
Therefore by using the formula we can write the RHS denominator as
⇒dudx+dudy+dudz=3(x2+y2+z2−xy−yz−zx)(x+y+z)(x2+y2+z2−xy−yz−xz)\Rightarrow \dfrac{{du}}{{dx}} + \dfrac{{du}}{{dy}} + \dfrac{{du}}{{dz}} = \dfrac{{3\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)}}{{\left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)}}
Therefore, the same term will cancel each other and we get
⇒dudx+dudy+dudz=3(x+y+z)\Rightarrow \dfrac{{du}}{{dx}} + \dfrac{{du}}{{dy}} + \dfrac{{du}}{{dz}} = \dfrac{3}{{\left( {x + y + z} \right)}}
Therefore, the equation we can write as

⇒(x+y+z)(ux+uy+uz)=3 \Rightarrow \left( {x + y + z} \right)\left( {{u_x} + {u_y} + {u_z}} \right) = 3

Additional information:
This type of question becomes very easy if we know the differentiation and some basic algebraic formula. Through this, we can easily solve it. As here in this question what we did is just the basic differentiation and by using one of the properties of algebra we get the values of the equation easily.

Note:
In this question, while doing the calculation we have to be aware and do the step-by-step solution as it will reduce the chances of error, and also we will not get confused while solving it. This question can also be asked in a different way as instead of ux,uy,uz{u_x},{u_y},{u_z}they will give us the question in the form of a derivative. So we don’t have to panic. The question is the same.