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Second Derivative Doesn't Work!

  1. Mar 16, 2005 #1
    Starting with y^2-xy=2, then y'=y/(2y-x)


    substitute y' in:
    y''=(2y-x)(y/2y-x) - y(2(y/2y-x)-1)

    the (2y-x) cancels:
    y''=(y) - y(2y/2y-x)-1)

    y'' = y - (2y^2)/(2y-x)-y

    Somewhere I need y^2-xy so I can replace it with the original 2, but I'm not getting it.
  2. jcsd
  3. Mar 16, 2005 #2
    No. He wants to implicitly differentiate to get [tex]\frac{d^2 y}{d x^2}[/tex].

    [tex]y^2 - xy = 2 \Longrightarrow 2y\frac{d y}{dx} - y - x\frac{dy}{dx} = 0 \Longrightarrow \frac{dy}{dx} = \frac{y}{2y-x}[/tex]

    [tex] \Longrightarrow \frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}}{2y-x} \ - \ \frac{y(2\frac{dy}{dx} - 1)}{(2y-x)^2} = \frac{\frac{y}{2y-x}}{2y-x} \ - \ \frac{y\left( \frac{2y}{2y-x} - 1\right)}{(2y-x)^2}[/tex]

    [tex]= \frac{y}{(2y-x)^2} \ - \ \frac{y\frac{2y - (2y-x)}{2y-x}}{(2y-x)^2} = \frac{y}{(2y-x)^2} \ - \ \frac{yx}{(2y-x)^3} = \frac{y(2y-x) - xy}{(2y-x)^3} = \frac{2y^2 - 2xy}{(2y-x)^3} = \frac{4}{(2y-x)^3}[/tex]
  4. Mar 16, 2005 #3
    I don't really follow. Why is the second derivative = first deriv - (top)(bottom)/(bottom)^2?

    In the book, the quotient rule is ((bottom)(top)' - (top)(bottom)')/(bottom)^2. That's what I tried to do, but it didn't work.
  5. Mar 16, 2005 #4


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    Data chose to differentiate by using the product rule on the function [tex]y(x)*\frac{1}{2y-x}[/tex]

    And, I have to add, the second derivative ALWAYS works, even if you don't know what to use it for..
    Last edited: Mar 16, 2005
  6. Mar 16, 2005 #5


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    Instead of looking for a y^2 - xy that you can replace with 2, maybe you could figure out what to replace y^2 with? Why'd you want to do that anyways?

    (I'm assuming your work was right)
  7. Mar 16, 2005 #6


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    That's not what you did. Look above. It should be y'/(2y-x)-y(2y'-1)/(2y-x)^2.
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