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An implicit problem

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Derivate [itex]y[/itex] with regards to [itex]x, \frac{dy}{dx}[/itex]

    [tex]2xy+y^2-4x=10[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I am not very good at differentiation, so I havent got off to a good start, really.
    I guess I can differentiate [itex]y^2 and -4x[/itex] "the usual way", but I am kind of stuck with the [itex]2xy[/itex]-part.

    I also recon I have to put the equation = 0
     
  2. jcsd
  3. Dec 9, 2011 #2

    SammyS

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    To find the derivative of 2xy, use the product rule.
     
  4. Dec 9, 2011 #3

    I like Serena

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    Hi Twinflower! :smile:

    How did you differentiate y2?
     
  5. Dec 9, 2011 #4
    So [itex](2x)'(y) + (2x)(y)' ---> y + 2x[/itex] ?

    That would yeld someting like: y + 2x +2y - 1, but the answer is supposed to be
    [tex]\frac{dy}{dx} = \frac{2-y}{x+y}[/tex]
     
  6. Dec 9, 2011 #5
    [itex](y^2)' = 2y[/itex]

    I think :)
     
  7. Dec 9, 2011 #6

    SammyS

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    (2x)' = 2, because x' = 1.

    Where did y' disappear to?
     
  8. Dec 9, 2011 #7

    I like Serena

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    I'm afraid you can't simplify (2x)(y)'.
    y' is what you want to find.


    Not quite.
    You should apply the chain rule here, and leave the y'.
     
  9. Dec 9, 2011 #8

    Mark44

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    No.
    d/dx(y2) = 2y[itex]\cdot[/itex]dy/dx = 2y[itex]\cdot[/itex]y'

    You didn't use the chain rule.
     
  10. Dec 9, 2011 #9
    Ok guys, I am gonna sit down and try to work this out.
    Thanks for pushing me in the right direction.

    I'll be back later :)
     
  11. Dec 9, 2011 #10
    OK, I finally got it:
    [tex]2xy+y^2-4x=10 [/tex]

    Differentiating each term:

    [tex]\frac{dy}{dx}2xy = (2x)' \times y + 2x \times y' = 2y +2x \frac{dy}{dx}[/tex]
    [tex]\frac{dy}{dx}y^2 = 2y \frac{dy}{dx}[/tex]
    [tex]\frac{dy}{dx}-4x = -4[/tex]
    [tex]\frac{dy}{dx} 10 = 0[/tex]

    Combining the terms yelds:

    [tex]2y +2x \frac{dy}{dx}+2y \frac{dy}{dx}-4=0[/tex]

    Diving each term by 2 and combining both y'-terms:

    [tex]\frac{dy}{dx}\times (x+y) + y-2 = 0[/tex]

    Subtracting (y-2) on both sides:

    [tex]\frac{dy}{dx}\times (x+y)=2-y[/tex]

    Dividing by (x+y) yelds the final answer:

    [tex]\frac{dx}{dy}=\frac{2-y}{x+y}[/tex]


    Edit: Alcubierre posted the answer while I was creating mine :)
     
    Last edited: Dec 9, 2011
  12. Dec 9, 2011 #11
    Haha we posted pretty much at the same time. Congratulations though, good job!
     
  13. Dec 9, 2011 #12

    I like Serena

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    Good! :smile:

    Btw, you did make a mistake here:
    [tex]\frac{dy}{dx}y^2 = 2y + \frac{dy}{dx}[/tex]
    which should be:
    [tex]\frac{dy}{dx}y^2 = 2y \cdot \frac{dy}{dx}[/tex]
    But apparently you fixed that while moving along.
     
  14. Dec 9, 2011 #13
    Ah, thanks Serena. Just a typo, it's not in my book.
    Gonna fix my post to avoid any confusion in case others are trying to use this thread :)
     
  15. Dec 9, 2011 #14

    I like Serena

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    Hey Twinflower.

    You still got 1 typo left, which is the one I actually mentioned. ;)
     
  16. Dec 9, 2011 #15
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