Question

Please, help me

Answer 1

Proof

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

mathematical physics

Answer 6

that could be in the later part of Mathematical Physics...... I only know the first half and have not done a proof like this

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

awesome @Michele_Laino , thank you

Answer 14

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

Answer 15

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.