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Homework Help: Implicit Differentiation: two different answers found. Please explain.

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Could anyone explain that I got two different answers for this question: find [tex]dy/dx[/tex] of [tex]\frac{x}{x+y}-\frac{y}{x}=4[/tex]

    2. Relevant equations



    3. The attempt at a solution

    1. using quotient rule:
    [tex]\frac{x+y-(1+dy/dx)x}{(x+y)^{2}}-\frac{x\frac{dy}{dx}-y}{x^{2}}=0[/tex]
    [tex]\frac{y}{(x+y)^{2}}-\frac{x}{(x+y)^{2}}\frac{dy}{dx}+\frac{y}{x^{2}}-\frac{1}{x}\frac{dy}{dx}=0[/tex]
    [tex](\frac{x}{(x+y)^{2}}+\frac{1}{x})\frac{dy}{dx}=\frac{y}{(x+y)^{2}}+\frac{y}{x^{2}}[/tex]
    [tex]x(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})\frac{dy}{dx}=y(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})[/tex]
    [tex]\frac{dy}{dx}=y/x[/tex]

    2. simplify using common denominator before taking derivative:
    [tex]\frac{x^{2}}{x^{2}+xy}-\frac{xy+y^{2}}{x^{2}+xy}=4[/tex]
    [tex]x^{2}-xy-y^{2}=4x^{2}+4xy[/tex]
    [tex]-y^{2}=3x^{2}+5xy[/tex]
    [tex]-2y\frac{dy}{dx}=6x+5y+5x\frac{dy}{dx}[/tex]
    [tex]\frac{dy}{dx}=-\frac{6x+5y}{2y+5x}[/tex]
     
  2. jcsd
  3. Mar 7, 2010 #2

    Dick

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    They aren't really different. Pick a point on the original curve like y=1, x=(sqrt(13)-5)/6. Substitute it into both of your expressions for dy/dx. You'll get the same thing. Coincidence?
     
  4. Mar 7, 2010 #3

    Dick

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    Or just try it with a braindead simple example. Take y/x=1. Do it one way and you get dy/dx=y/x, do it the other way and you get dy/dx=1. Which one is right?
     
  5. Mar 7, 2010 #4
    I have tried a few examples, and they all came out to be the same. Wow, I am so amazed that two equations that look so different can be the same thing.
    I also tried to let the two equations equal:
    1. [tex]
    \frac{dy}{dx}=y/x
    [/tex]
    2. [tex]
    \frac{dy}{dx}=-\frac{6x+5y}{2y+5x}
    [/tex]
    Simplifying it, I got
    [tex]y/x= -\frac{6x+5y}{2y+5x}[/tex]
    [tex]2y^{2}+5xy=-6x^{2}-5xy[/tex]
    [tex]6x^{2}+10xy+2y^{2}=0[/tex]
    [tex]3x^{2}+5xy+y^{2}=0[/tex]

    Which is the same as the original equation, shown in the process using the second method, the one says "simplifying using common denominator".
     
  6. Mar 7, 2010 #5
    Ya, thank you so much. What I had always been doing were trying disprove the first method, the one saying "using quotient rule", because of the step following this:
    [tex]
    x(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})\frac{dy}{dx }=y(\frac{1}{(x+y)^{2}}+\frac{1}{x^{2}})
    [/tex]
     
  7. Mar 7, 2010 #6

    Dick

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    Just to be clear. y/x and -(6x+5y)/(2y+5x) aren't the same expression. But you can prove they are equal if you combine them with x/(x+y)-y/x=4. Did you see post 3? It's a really simple example of the same thing.
     
  8. Mar 7, 2010 #7
    Ya, I know they are not the same thing. They are just related to the original equation.

    By post three, you mean this?
    I didn't understand it. Were you trying to tell me that all three of these equations relate to each other?
     
  9. Mar 7, 2010 #8

    Dick

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    Ok, (y/x)=1. (y/x)'=y'/x-y/x^2=0. Solve for y'=y/x. Or reduce the original equation and get y=x. So y'=1. y'=1 and y'=y/x LOOK different. Until you remember the original equation was y/x=1. It's the same sort of difference you are looking at. Just a lot more obvious.
     
  10. Mar 7, 2010 #9
    I was confused about what you meant by "do it". I thought you wanted me to plug that equation into the ones I was doing. Now it's clear. Thanks a lot.
     
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