I'm trying to solve a question in which it is given that u=f(x,y,z) and it is asked to find (∂/∂x+∂/∂y+∂/∂z)^2 of u . How do I solve this? Should I apply the formula (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca ? Or any other easy method ?
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Is there more information? Because right now it sounds like Laplace or Poisson equation depending on the conditions – Triatticus Aug 28 '15 at 07:30
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Dex, well the actual question is If u=log(x^3+y^3+z^3-3xyz),show that (∂/∂x+∂/∂y+∂/∂z)^2 of u = (-9)/(x+y+z)^2 . – Aug 28 '15 at 07:37
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Oh so the question is to take partial derivatives? – Triatticus Aug 28 '15 at 07:38
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Yes Dex. Anyway thanks, it has been answered. :) – Aug 28 '15 at 08:01
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This is merely a suggestion on how to book keep your calculation. I will leave the actual calculations to you.
I suggest that you do it in two steps. First, calculate $$ h=\Bigl(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z}\Bigr)u=\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}. $$ (Note, that you by symmetry only need to do one of the differentiations.) After simplification, I get $$ h=\frac{3}{x+y+z}. $$ Then, calculate $$ \Bigl(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z}\Bigr)^2u=\Bigl(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z}\Bigr)h=\frac{\partial h}{\partial x}+\frac{\partial h}{\partial y}+\frac{\partial h}{\partial z}. $$ When I do the calculation, I get the result $$ -\frac{9}{(x+y+z)^2}. $$
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