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    [\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!!!!
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    what a waste of time this was!
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    one massively important note I forgot to add in, this is vector calculus, the i term is -y and the j term is x, hence when finding the directional derivative, it becomes the partial der. wrt x of the i term + the partial der. wrt y of the j term.
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    admitting defeat on this colossal mess, going to use the quotiant rule (shiver) instead.

    If anyone sees this, could you please explain how I would use a combination of the chain and product rule to solve the two partial derivatives that come about when I try to work out the divergence??
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    man got aired
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    (Original post by Et Tu, Brute?)
    [\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}
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    (Original post by ghostwalker)
    -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:



    \frac{2xy}{(x^2 + y^2)} - \[frac{2xy}{(x^2 + y^2)}
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    (Original post by Et Tu, Brute?)
    ok, so this even more annoying, as I had this initially, however, the answer has no [x^2 + y^2]^2 denominator!

    they have:



    \frac{2xy}{(x^2 + y^2)} - \[frac{2xy}{(x^2 + y^2)}

    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.
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    (Original post by ghostwalker)
    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.
    yes, after getting div u it is nothing more than normal partials after that as it is no longer a vector.

    Partial wrt x of the -y/a and partial wrt y of x/a, where a = x^2+y^2 (just for clarity here)


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


    The answer is actually zero, as I'm sure you might have noticed, so they've maybe got caught up in that and forgot to include the ^2 in the denominator maybe.
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    (Original post by Et Tu, Brute?)
    So 100%, the denominator becomes (x^2+y^2)^2 ???
    Yes. Doing it via the quotient rule yields the same result.

    And here's Wolfram's attempt.
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    (Original post by Et Tu, Brute?)
    [\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.
    Spoiler:
    Show
    hint: \frac{d}{dx}a(x^n+b)^c = acnx^{n-1}(x^n + b)^{(c-1)}
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    (Original post by natninja)
    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.
    Spoiler:
    Show
    hint: \frac{d}{dx}a(x^n+b)^c = acnx^{n-1}(x^n + b)^{(c-1)}

    yeah I did this initially and ended up with the correct answer, though they have the incorrect answer in the answer sheet, hence why I ended up trying to do other stuff. Should have just trusted myself and wolfram alpha'd it before to double check.

    thanks for the help, sorry for wasting time
 
 
 
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