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Xy + x - 2y -1 = 0 Help

  1. Oct 16, 2009 #1
    Bought Schaum's Calculus 5th edition to clean up my calculus worries but I haven't cleared this one up yet;

    [tex] xy + x - 2y -1 = 0 [/tex]

    [tex] Solving for (y); [/tex]
    [tex] xy - 2y = - x + 1 [/tex]
    [tex] y(x - 2) = - x + 1 [/tex]
    [tex] y = \frac {-x+1}{x-2} [/tex]

    [tex]Taking the derivative of y = \frac {-x+1}{x-2} using the quotient rule[/tex]

    [tex] \frac {dy}{dx} = \frac {- 1 (x - 2) - 1 (- x + 1)} {(x - 2)^2} [/tex]

    [tex] \frac {dy}{dx} = \frac {- x + 2 + x - 1)} {(x - 2)^2} [/tex]

    [tex] \frac {dy}{dx} = \frac {1} {(x - 2)^2} [/tex]

    Now, my problem is that when I do this calculation via Implicit Differentiation, this is what I get;

    [tex] xy + x - 2y -1 = 0 [/tex]

    [tex] y + x\frac {dy}{dx} + 1 - 2\frac {dy}{dx} = 0 [/tex]
    [tex] x\frac {dy}{dx}- 2\frac {dy}{dx} = - y - 1 [/tex]
    [tex] \frac {dy}{dx}( x - 2) = - y - 1 [/tex]
    [tex] \frac {dy}{dx} = \frac {- y - 1} {x - 2} [/tex]

    So;
    1.My two answers don't agree, aren't they supposed to? If not, why don't they?

    2.My book gives a different answer which I cannot get,
    [tex] \frac{1+ y} {2 - x} [/tex]
    Am I careless with signs or is my book wrong?

    Gratias tibi ago !
     
  2. jcsd
  3. Oct 16, 2009 #2

    lanedance

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    i haven't checked all your working yet but the main differnce is the y in the implictly derived equation if you sub in for y and have done everything correctly they will be the same
     
  4. Oct 16, 2009 #3

    lanedance

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    [tex] xy + x - 2y -1 = 0 [/tex]
    differntiating
    [tex] y + xy' + 1 - 2y'= 0 [/tex]
    re-arranging
    [tex] y'(x-2)= -(y+1) [/tex]
    so
    [tex] y'= \frac{(y+1)}{2-x} [/tex]

    your answer is in fact the same as the book and above, try multiplying through by (-1)/(-1)
     
  5. Oct 16, 2009 #4
    That's great, I got very worried I was still getting wrong answers (but also excited at the thought I found another wrong answer in a book lol) But is the quotient rule derivation wrong? It seems very plausible the way I worked it out yet there's a different answer. How so?
     
  6. Oct 16, 2009 #5
    They're both the same thing. You've got y=(1-x)/(x-2) So,

    y+1=(1-x)/(x-2)+1=(1-x+x-2)/(x-2)=1/(2-x) So

    (y+1)/(2-x)=1/(2-x).^2
     
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