Vector and partial derivatives

[Crazy Gnome]

Homework Statement

Two charges one located at P at the position (x,y,z) and P' at the position (x',y',z')

Let f= 1/R.
Calculate Fx= partial derivative of f with respect to x.
Calculate Fx'= partial derivative of f with respect to x'.

There are sub question involving the same thing with other variables but if you could help me figure out the first part that should go easy.

The Attempt at a Solution

I solved R to be = (x-x')i + (y-y')j + (z-z')k... i think that's right but I do not know how to go about finding the partial derivatives of f. I have never really worked with vectors let alone inverse vectors with calculus.

 

[Dick]

There is no such thing as an 'inverse vector'. I would guess they mean R=|(x,y,z)-(x',y',z')|, Which is R=sqrt((x-x')^2+(y-y')^2+(z-z')^2).

 

[Crazy Gnome]

There is no such thing as an 'inverse vector'. I would guess they mean R=|(x,y,z)-(x',y',z')|, Which is R=sqrt((x-x')^2+(y-y')^2+(z-z')^2).

Right, I got that part. But what is a partial derivative of 1/ a vector?

 

[Dick]

It can't be a vector. 1/'a vector' doesn't make any sense. Are you sure they didn't write 1/|R|? It's just the partial derivative of a scalar function.

 

[Crazy Gnome]

It can't be a vector. 1/'a vector' doesn't make any sense. Are you sure they didn't write 1/|R|? It's just the partial derivative of a scalar function.

Right *bangs head against the wall*. The only difference on the homework sheet between the vector and the magnitude is the vector is in bold. My stupid mistake.


-Thanks