How to do I this disgusting partial derivative?

$[\frac{-yi+xj)}{x^2 + y^2} \ \$
ok, so no more latex for you guys, blame my lack of experience as well as the awful implementation of it on this site, never had such difficulty with it before this site, joke!!!


frac{-y} {x^2 + y^2}\) + frac{x} {x^2 + y^2}

side note, this is the worst site in the world for latex!!!!!!!!!!!

product rule followed by chain rule here I think, but to be honest I don't recall having to do something quite as complicated as this in a long time, so I need some guidance.

First attempt, I got rid of the denominator by making it $-y(x^2 + y^2)^{-1}$, I then made this $-(yx^2 + y^3)^{-1}$ and use chain rule to give $2x+3y^2(x^2 + y^2)^{-2}$(cba fixing inside bracket, waste of time) this is so very wrong (though I don't see why I can't multiply out the bracket and use the chain rule, ie avoid the product rule!).


Then I tried product rule first, though this is giving me a massive headache!

****ing chirst what a mess!!!!

$-y(x^2 + y^2)^{-1}\not=-(yx^2 + y^3)^{-1}$ because of the -1 exponent.

Rather $-y(x^2 + y^2)^{-1}=-(\frac{x^2}{y} + y)^{-1}$ which is not terrribly helpful.

I'd go with the just the chain rule - remember y is treated as a constant:

$\displaystyle\frac{\partial }{\partial x}\left(-y(x^2 + y^2)^{-1}\right)$

$=(-y)(-1)(2x)(x^2 + y^2)^{-2}$

$=2xy(x^2 + y^2)^{-2}$

ok, so this even more annoying, as I had this initially, however, the answer has no [x^2 + y^2]^2 denominator!

they have:

I don't claim any proficiency in vector calculus, but if all they're doing is taking the partial derivative wrt x, then I would say "they" are in error.

Edit: Your post is changing again. I'll await the final version.

So 100%, the denominator becomes (x^2+y^2)^2 ???

$[\frac{-yi+xj)}{x^2 + y^2} \ \$
ok, so no more latex for you guys, blame my lack of experience as well as the awful implementation of it on this site, never had such difficulty with it before this site, joke!!!


frac{-y} {x^2 + y^2}\) + frac{x} {x^2 + y^2}

side note, this is the worst site in the world for latex!!!!!!!!!!!

product rule followed by chain rule here I think, but to be honest I don't recall having to do something quite as complicated as this in a long time, so I need some guidance.

First attempt, I got rid of the denominator by making it $-y(x^2 + y^2)^{-1}$, I then made this $-(yx^2 + y^3)^{-1}$ and use chain rule to give $2x+3y^2(x^2 + y^2)^{-2}$(cba fixing inside bracket, waste of time) this is so very wrong (though I don't see why I can't multiply out the bracket and use the chain rule, ie avoid the product rule!).


Then I tried product rule first, though this is giving me a massive headache!

****ing chirst what a mess!!!!

So we have a vector

$(-y(x^2 + y^2)^{-1}, x(x^2 + y^2)^{-1}, 0)$

$\bigtriangledown .(-y(x^2 + y^2)^{-1}, x(x^2 + y^2)^{-1}, 0) = \frac{\partial}{\partial x}(-y(x^2 + y^2)^{-1}) + \frac{\partial}{\partial y} (x(x^2 + y^2)^{-1})$

and it becomes a pretty simple example of the chain rule. Spoilertagged a hint.