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Math Help - Is my answer for derivative right

  1. #1
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    Is my answer for derivative right

    Find derivative of y= (x)/(x+y)

    I used the quotient rule and then got y' on one side. My final answer is:
    [(x+y) - y] / (-2)


    Is that correct?
    Thanks!
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  2. #2
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    y = \frac {x}{x + y}

    y(x + y) - x = 0

    xy + y^2 - x = 0

    \frac {dy}{dx} = \frac {1 - y}{x + 2y}
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  3. #3
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    Quote Originally Posted by derfleurer View Post
    y = \frac {x}{x + y}

    y(x + y) - x = 0

    xy + y^2 - x = 0

    \frac {dy}{dx} = \frac {1 - y}{x + 2y}

    Now I'm confused because if you do it this way you get one answer and if you leave it as a quotient, I get a different answer.... y/[x^2 + 2xy +y^2 +x]

    I can't figure out why.
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  4. #4
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    The answer that he/she gave looks correct to me. Please show your work for using the quotient rule and I'll try to spot an error if there is one.
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  5. #5
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    Quote Originally Posted by Jameson View Post
    The answer that he/she gave looks correct to me. Please show your work for using the quotient rule and I'll try to spot an error if there is one.
    Ok here's one of the ways I did it:

    y' = (x+y)(1) - (x)(1+y') / [(x+y)^2]
    = x+y - x - xy' / [(x+y)^2]
    =y-xy' / [(x+y)^2]
    y'(x+y)^2 = y-xy'
    y'(x+y)^2 +xy' = y
    y'[(x+y)^2 +y = y
    y' = y / [(x+y)^2 +x)
    y' = y / [x^2 + 2xy +y^2 +x]
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  6. #6
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    Your work looks fine to me. I have a feeling that both answers are equivalent, but the algebra to change your form to the other is kind of messy. Maybe someone else can show that they are equal.
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  7. #7
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    Looks right. Same thing I got using quotient rule. Wouldn't dare delve into the math to simplify them.
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