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Use implicit differentiation to find dy/dx [Answer check]

  1. Sep 13, 2008 #1
    Use implicit differentiation to find [tex]\frac{dy}{dx}[/tex] for xy[tex]^{2}[/tex] – yx[tex]^{2}[/tex] = 3xy

    i've answered the question but i think i'm doing it wrong
    any help is appreciated!

    x(2y)[tex]\frac{dy}{dx}[/tex] – y(2x) = 3xy
    2xy [tex]\frac{dy}{dx}[/tex] – 2yx = 3xy
    2xy[tex]\frac{dy}{dx}[/tex] = 5xy
    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{5xy}{2xy}[/tex]
    [tex]\frac{dy}{dx}[/tex] = 3xy

    Thank you!
     
  2. jcsd
  3. Sep 13, 2008 #2

    Defennder

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    Your first line is the mistake. You forgot to differentiate the other term on the left.
     
  4. Sep 13, 2008 #3
    Ok i'm not sure if i've done any better this time but here it goes:

    x(2y)[tex]\frac{dy}{dx}[/tex] - y(2x)[tex]\frac{dy}{dx}[/tex] = 3xy

    2xy[tex]\frac{dy}{dx}[/tex] - 2yx[tex]\frac{dy}{dx}[/tex] = 3xy

    [tex]\frac{dy}{dx}[/tex] (2xy - 2yx) = 3xy

    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{3xy}{2xy - 2yx}[/tex]

    i'm not sure if the 2xy and 2yx would cancel each other out but it made more sense then dividing by a zero!!
    thank you
     
  5. Sep 13, 2008 #4

    danago

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    You didnt take Defenner's advice; When you differentiate one side of the equals sign, you need to also differentiate the other side. So you need to differentiate the 3xy part too.

    Also, be careful how you are differentiating xy2 for example. You need to remember that xy2 is the product of two functions, x and y2. You need to use the product rule.
     
  6. Sep 13, 2008 #5
    Alright...i THINK i got it...lets see

    y[tex]^{2}[/tex] + x(2y) [tex]\frac{dy}{dx}[/tex] - x[tex]^{2}[/tex] [tex]\frac{dy}{dx}[/tex] - y(2x) = 3y + 3x [tex]\frac{dy}{dx}[/tex]

    x(2y)[tex]\frac{dy}{dx}[/tex] - x[tex]^{2}[/tex] [tex]\frac{dy}{dx}[/tex] - 3x [tex]\frac{dy}{dx}[/tex] = 3y - y[tex]^{2}[/tex] + y(2x)

    [tex]\frac{dy}{dx}[/tex] (2yx - x[tex]^{2}[/tex] - 3x) = 3y - y[tex]^{2}[/tex] + 2xy

    [tex]\frac{dy}{dx}[/tex] = (3y - y^2) / (-x^2 - 3x)
    am I getting any closer to a correct answer?
     
    Last edited: Sep 13, 2008
  7. Sep 13, 2008 #6

    Dick

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    You are getting there. But you can't 'cancel' the 2xy from numerator and denominator, can you? Back up one line and try again.
     
  8. Sep 13, 2008 #7

    danago

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    Where did the 2xy from both the numerator and denominator go in the last step? Other than that, looks alright to me.

    Edit: Looks like Dick beat me to it :p
     
  9. Sep 13, 2008 #8
    Ok great! i'll pop them back into the equation!!
    Thank you guys!!!
     
  10. Sep 13, 2008 #9

    Dick

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    Fine. As long as you understand why you can't cancel that way and promise not to do it again. (1+2)/(3+2)=3/5, but it turns into 1/3 if you 'cancel' the 2.
     
  11. Sep 13, 2008 #10
    I could not read your picture, so perhaps this is superfluous.

    xy2 - x2y - 3xy = 0
    dx/dx = 1
    d/dx(xy2 - x2y - 3xy) = d0/dx = 0
    y2 + 2xy(dy/dx) - 2xy - x2(dy/dx) - 3y - 3x(dy/dx)=0
    y2 - 2xy - 3y = x2(dy/dx) + 3x(dy/dx) - 2xy(dy/dx)
    dy/dx = (y2 - 2xy - 3y)/(x2 + 3x - 2xy)
     
  12. Sep 13, 2008 #11

    I understand :biggrin: Thank you I promise I will try really hard not to do it again!
    Take care =)
     
  13. Jun 24, 2010 #12

    u forgot the use of product rule here.....and therefore.....your answer is wrong....!!
     
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