anonymous
  • anonymous
Please, help me
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Proof \[\frac{ \textbf{x }- \textbf{x}_i }{ \left| \textbf{x} - {\textbf{x}_i} \right|^3 } = \bigtriangledown \left( \frac{ 1 }{ x-x' } \right)\]
TheCatMan
  • TheCatMan
i say to find the area of the triangle using other info to the question (if there is none i have no clue) distribute to the right side of the equation and cross eliminate to solve
UsukiDoll
  • UsukiDoll
WHAT THE! SOlving has nothing to do with writing a complete proof. And I don't know which subject this came from. It could be from some grad school Math course :S

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UsukiDoll
  • UsukiDoll
I know that for proofs like this you either start at the left apply some theorems...definitions lalalalalla and achieve the right or start at the right apply some theorems...definitions lalalla and achieve the left.
anonymous
  • anonymous
mathematical physics
UsukiDoll
  • UsukiDoll
that could be in the later part of Mathematical Physics...... I only know the first half and have not done a proof like this
TheCatMan
  • TheCatMan
i wish i could help but im dumber than a fart in a crowded room.
alekos
  • alekos
it really means nothing without some sort of context. tel us more about the problem
Michele_Laino
  • Michele_Laino
here are some steps: we can write: \[{\mathbf{x}} - {{\mathbf{x}}_i} = {\mathbf{r}}\] where: \[{\mathbf{r}} \cdot {\mathbf{r}} = {x^2} + {y^2} + {z^2}\]
Michele_Laino
  • Michele_Laino
so we have: \[\large \begin{gathered} \frac{\partial }{{\partial x}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial x}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{x}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial y}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial y}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{y}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial z}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial z}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{z}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
\[\begin{gathered} \frac{\partial }{{\partial x}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial x}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{x}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial y}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial y}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{y}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial z}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial z}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{z}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
and finally: \[\Large \nabla \left( {\frac{1}{r}} \right) = - \frac{{\mathbf{r}}}{{{r^3}}}\]
anonymous
  • anonymous
awesome @Michele_Laino , thank you
anonymous
  • anonymous
i also thank to @TheCatMan , @UsukiDoll , @alekos
UsukiDoll
  • UsukiDoll
what @Michele_Laino latexed is similar to an example from a book I had to use almost a year ago. There was partial derivatives, product rule, and chain rule involved.

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