If u=log left(x^3+y^3+z^3-3 x y zright), then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given, $u=\log \left(x^3+y^3+z^3-3 x y z\right)$.We know that $x^3+y^3+z^3-3 x y z = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.So, $u = \log(x+y+z) + \log(x^

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given, $u=\log \left(x^3+y^3+z^3-3 x y z\right)$.We know that $x^3+y^3+z^3-3 x y z = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.So, $u = \log(x+y+z) + \log(x^2+y^2+z^2-xy-yz-zx)$.Now, calculate the partial derivatives:$u_x = \frac{\partial u}{\partial x} = \frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$,$u_y = \frac{\partial u}{\partial y} = \frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$,$u_z = \frac{\partial u}{\partial z} = \frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$.Summing these up:$u_x+u_y+u_z = \frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)} = \frac{3}{x+y+z}$.Therefore, $(x+y+z)(u_x+u_y+u_z) = 3$.

If $u=\log \left(x^3+y^3+z^3-3 x y z\right)$, then $(x+y+z)(u_x+u_y+u_z)$ is equal to

Similar Questions

If two distinct chords drawn from the point $(p, q)$ to the circle $x^{2} + y^{2} = px + qy$ (where $pq \neq 0$) are bisected by the $x$-axis,then:

$Na_2S + Na_2[Fe(CN)_5NO] \longrightarrow Na_4[Fe(CN)_5NOS]$; what are the oxidation numbers of $Fe$ in the reactant complex and the product complex,respectively?

What is the product obtained in the following reaction?

The coefficient of $x^5$ in the expansion of $(1+x^2)^5(1+x)^4$ is:

Among the following,which is water insoluble?

Vedclass Test Series

Exam Paper Generator

Online Exam Module