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Math Help - Partial integration - correct solutions?

  1. #1
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    Partial integration - correct solutions?

    Let f(x; y; z) = (x^3+3xy^2-3xyz-z^4)/(x^2 + z^2)

    Calculate

    a. @f/@x

    b. @f/@y

    c. @f/@z

    I suppose that the solution for the partial derivative of the function to x is too simple as I get: 3(x^2+y^2-yz)/2x

    Am I right when I have to use the quotient rule? as it is fg - gf / g^2......But what is then g only x^2 or x^2+z^2 and is f (x^3+3xy^2-3xyz-z^4) or just (x^3+3xy^2-3xyz) (as the unused variables are held constant?

    so is this approach correct:

    (3x^2+3y^2-3yz)*(x^2) - 2x(x^3+3xy^2-3xyz) / x^4

    thank you in advance....
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  2. #2
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    Re: Partial integration - correct solutions?

    Quote Originally Posted by FireTheft View Post
    Let f(x; y; z) =
    Calculate

    a. @f/@x

    so is this approach correct:

    (3x^2+3y^2-3yz)*(x^2) - 2x(x^3+3xy^2-3xyz) / x^4

    thank you in advance....
    Your last expression is close to correct. I suspect that notation is getting in the way as well. We already have a function f(x, y, x) so using it in the quotient rule is a little confusing. So I am going to say
    f(x, y, z) = \frac{ x^3+3xy^2-3xyz-z^4}{x^2 + z^2}

    And I am going to define the quotient rule as
    given a function
    u(x) = \frac{m(x)}{n(x)}

    then
    u'(x) = \frac{m'(x)\cdot n(x) - m(x) \cdot n'(x)}{(n(x))^2}

    So to get the derivative then choose m(x, y, z) = x^3+3xy^2-3xyz-z^4 and n(x) = x^2 + z^2

    See what you can do with that.

    -Dan
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  3. #3
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    Re: Partial integration - correct solutions??

    So am I right that u`(x) equals:

    (3x^2+3y^2-3yz)*(x^2+z^2) - 2x(x^3+3xy^2-3xyz-z^4) / (x^2 + z^2)^2

    so is the correct answer for u(x):

    (x^4+3xy^2+3x^2yz+3x^2z^2+3y^2z^2-3yz^3-6x^3y^2+2xz^4)/(x^2+z^2)^2

    for u(y) I got:

    (6x^3y-3x^3z+6xyz^2-3xz^3)/(x^2+z^2)^2

    for u`(z) I got:

    (-3x^3y-4x^2z^3-3xyz^2-2z^5-2x^3z-6xy^2z+6xyz)/(x^2+z^2)^2

    I hope someone can confirm these solutions ;-)

    Thx a lot
    Last edited by FireTheft; September 18th 2012 at 05:40 AM.
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