Partial Differentiation: If v = log(x^3+y^3+z^3-3xyz), show (∂/∂x + ∂/∂y + ∂/∂z)v =

[Just]

Partial Differentiation: If v = log(x^3+y^3+z^3-3xyz), show (∂/∂x + ∂/∂y + ∂/∂z)v =

10. If \(\displaystyle v\, =\, \log\left(x^3\, +\, y^3\, +\, z^3\, -\, 3xyx\right),\) show that

. . . . .\(\displaystyle \left(\dfrac{\partial}{\partial x}\, +\, \dfrac{\partial}{\partial y}\, +\, \dfrac{\partial}{\partial z}\right)^2\, v\, =\, \dfrac{-9}{(x\, +\, y\, +\, z)^2}\)

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I am not sure whether (∂/∂x + ∂/∂y + ∂/∂z)²v means ∂v²/∂²x + ∂v²/∂²y + ∂v²/∂²z or (∂v/∂x + ∂v/∂y + ∂v/∂z)²

I tried to solve it both the ways but didn't reach the required answer.

 

[Deleted member 4993]

10. If \(\displaystyle v\, =\, \log\left(x^3\, +\, y^3\, +\, z^3\, -\, 3xyx\right),\) show that

. . . . .\(\displaystyle \left(\dfrac{\partial}{\partial x}\, +\, \dfrac{\partial}{\partial y}\, +\, \dfrac{\partial}{\partial z}\right)^2\, v\, =\, \dfrac{-9}{(x\, +\, y\, +\, z)^2}\)

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I am not sure whether (∂/∂x + ∂/∂y + ∂/∂z)²v means ∂v²/∂²x + ∂v²/∂²y + ∂v²/∂²z or (∂v/∂x + ∂v/∂y + ∂v/∂z)²

I tried to solve it both the ways but didn't reach the required answer.

Please share your work with us - so that we assist you properly.

 

[HallsofIvy]

10. If \(\displaystyle v\, =\, \log\left(x^3\, +\, y^3\, +\, z^3\, -\, 3xyx\right),\) show that

. . . . .\(\displaystyle \left(\dfrac{\partial}{\partial x}\, +\, \dfrac{\partial}{\partial y}\, +\, \dfrac{\partial}{\partial z}\right)^2\, v\, =\, \dfrac{-9}{(x\, +\, y\, +\, z)^2}\)

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I am not sure whether (∂/∂x + ∂/∂y + ∂/∂z)²v means ∂v²/∂²x + ∂v²/∂²y + ∂v²/∂²z...

You should not even have considered that because \(\displaystyle (a+b+ c)^2\) is NOT equal to \(\displaystyle a^2+ b^2+ c^2\)!

] or (∂v/∂x + ∂v/∂y + ∂v/∂z)²

No, it is not that either. \(\displaystyle \left(\frac{\partial }{\partial x}+ \frac{\partial }{\partial y}+ \frac{\partial }{\partial z}\right)^2v\) means "apply the differential operator \(\displaystyle \frac{\partial}{\partial x}+ \frac{\partial}{\partial y}+ \frac{\partial}{\partial z}\) to v twice".

With \(\displaystyle v= ln(x^3+ y^3+ z^3- 3xyz)\), the derivative with respect to x, y, or z is \(\displaystyle \frac{1}{x^3+ y^3+ z^3- 3xyz}\) times the derivative of \(\displaystyle x^3+ y^3+ z^3- 3xyz\) with respect to x, y, or z, respectively.

\(\displaystyle \frac{\partial v}{\partial x}= \frac{3x^2- 3yz}{x^3+ y^3+ z^3- 3xyz}\). Calculate \(\displaystyle \frac{\partial v}{\partial y}\) and \(\displaystyle \frac{\partial v}{\partial z}\) and add them. Then do it again.

I tried to solve it both the ways but didn't reach the required answer.

 

[Just]

Thanks for the help.
So, I have reached to this.

∂v/∂x + ∂v/∂y + ∂v/∂z = 3(x²+y²+z²-xy-yz-xz)/(x³+y³+z³-3xyz)

And now I have no clue what to do.

 

[Just]

OK, so I partially differentiated it with respect to x, y and z separately. Then I added them together. What is the next step?