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Math Help - partial dereivatives

  1. #1
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    partial dereivatives

    f(x,y)=(x+y)/(x^2+2y^2+6)

    find the stationary points.

    i'm having trouble finding the partial derivatives. they came out really messy.
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  2. #2
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    Hello,

    Quote Originally Posted by dinNA89 View Post
    f(x,y)=(x+y)/(x^2+2y^2+6)

    find the stationary points.

    i'm having trouble finding the partial derivatives. they came out really messy.
    For example, let's take the one with respect to x.

    \frac{\partial (x+y)}{\partial x}=1

    \frac{\partial (x^2+2y^2+6)}{\partial x}=2x

    ---> \begin{aligned} \frac{\partial f(x,y)}{\partial x} & =\frac{(x^2+2y^2+6)-2x \cdot (x+y)}{(x^2+2y^2+6)^2} \\<br />
& =\frac{x^2+2y^2+6-2x^2-2xy}{(x^2+2y^2+6)^2} \\<br />
& =\frac{-x^2-2xy+2y^2}{(x^2+2y^2+6)^2} \end{aligned}
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  3. #3
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    Quote Originally Posted by dinNA89 View Post
    f(x,y)=(x+y)/(x^2+2y^2+6)

    find the stationary points.

    i'm having trouble finding the partial derivatives. they came out really messy.
    Not much mess that I can see. Just use the quotient rule:


    \frac{\partial f}{\partial x} = \frac{(x^2 + 2y^2 + 6) - 2x(x+y)}{\text{stuff}} = \frac{-x^2 -2xy + 2y^2 + 6}{\text{stuff}}


    \frac{\partial f}{\partial y} = \frac{(x^2 + 2y^2 + 6) - 4y(x+y)}{\text{stuff}} = \frac{x^2 -4xy - 2y^2 + 6}{\text{stuff}}


    I haven't worried about the denominators because you're gonna equate each of the partial derivatives to zero so it's only the numerators that matter.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    \frac{\partial f}{\partial x} = \frac{(x^2 + 2y^2 + 6) - 2x(x+y)}{\text{stuff}} = \frac{-x^2 -2xy + 2y^2 + 6}{\text{stuff}}


    \frac{\partial f}{\partial y} = \frac{(x^2 + 2y^2 + 6) - 4y(x+y)}{\text{stuff}} = \frac{x^2 -4xy - 2y^2 + 6}{\text{stuff}}
    now i'm having trouble solving the numerators to equal 0.
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  5. #5
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    Quote Originally Posted by dinNA89 View Post
    now i'm having trouble solving the numerators to equal 0.
    You have two equations. Add them together and simplify. You get xy = 2. Now substitute y = 2/x, say, into either one of the two equations and solve for x etc.
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