## anonymous one year ago Please, help me

1. anonymous

Proof $\frac{ \textbf{x }- \textbf{x}_i }{ \left| \textbf{x} - {\textbf{x}_i} \right|^3 } = \bigtriangledown \left( \frac{ 1 }{ x-x' } \right)$

2. 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

3. 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

4. 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.

5. anonymous

mathematical physics

6. 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

7. TheCatMan

i wish i could help but im dumber than a fart in a crowded room.

8. alekos

it really means nothing without some sort of context. tel us more about the problem

9. Michele_Laino

here are some steps: we can write: ${\mathbf{x}} - {{\mathbf{x}}_i} = {\mathbf{r}}$ where: ${\mathbf{r}} \cdot {\mathbf{r}} = {x^2} + {y^2} + {z^2}$

10. Michele_Laino

so we have: $\large \begin{gathered} \frac{\partial }{{\partial x}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial x}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{x}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial y}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial y}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{y}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial z}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial z}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{z}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \end{gathered}$

11. Michele_Laino

$\begin{gathered} \frac{\partial }{{\partial x}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial x}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{x}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial y}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial y}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{y}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \hfill \\ \frac{\partial }{{\partial z}}\left( {\frac{1}{r}} \right) = \frac{\partial }{{\partial z}}\left( {\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) = - \frac{z}{{{{\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)}^3}}} \hfill \\ \end{gathered}$

12. Michele_Laino

and finally: $\Large \nabla \left( {\frac{1}{r}} \right) = - \frac{{\mathbf{r}}}{{{r^3}}}$

13. anonymous

awesome @Michele_Laino , thank you

14. anonymous

i also thank to @TheCatMan , @UsukiDoll , @alekos

15. 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.