Question

If u = x²(y-z) + y²(z-x)+z²(x-y) then prove that du/dx + du/dy +du/dz = 0
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Answer 1

Answer:

here's the answer

Step-by-step explanation:

hope it help....

Answer 2

Solution:

Given that:

$\rm\longrightarrow u=x^{2}(y-z) + y^{2}(z - x) + z^{2}(x-y)$

Partial differentiating both sides with respect to x, we get:

$\rm\longrightarrow \dfrac{\partial u}{\partial x}=\dfrac{\partial}{\partial x}\{x^{2}(y-z)\} +\dfrac{\partial}{\partial x}\{y^{2}(z - x)\} +\dfrac{\partial}{\partial x}\{z^{2}(x-y)\}$

$\rm\longrightarrow \dfrac{\partial u}{\partial x}=2x(y-z)-y^{2}+z^{2} -(i)$

Partial differentiating both sides with respect to y, we get:

$\rm\longrightarrow \dfrac{\partial u}{\partial y}=\dfrac{\partial}{\partial y}\{x^{2}(y-z)\} +\dfrac{\partial}{\partial y}\{y^{2}(z - x)\} +\dfrac{\partial}{\partial y}\{z^{2}(x-y)\}$

$\rm\longrightarrow \dfrac{\partial u}{\partial y}=x^{2}+2y(z-x)-z^{2}-(ii)$

Partial differentiating both sides with respect to z, we get:

$\rm\longrightarrow \dfrac{\partial u}{\partial z}=\dfrac{\partial}{\partial z}\{x^{2}(y-z)\} +\dfrac{\partial}{\partial z}\{y^{2}(z - x)\} +\dfrac{\partial}{\partial z}\{z^{2}(x-y)\}$

$\rm\longrightarrow \dfrac{\partial u}{\partial z}=-x^{2}+y^{2} +2z(x-y)-(iii)$

Adding equations (i), (ii) and (iii), we get:

$\rm\longrightarrow \dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}=2x(y-z)-y^{2}+z^{2}+x^{2}+2y(z-x)-z^{2}-x^{2}+y^{2}+2z(x-y)$

$\rm\longrightarrow \dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}=2x(y-z)+2y(z-x)+2z(x-y)$

$\rm\longrightarrow \dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}=2xy-2xz+2yz-2yx+2zx-2zy$

$\rm\longrightarrow \dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}=0$

Hence Proved..!