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Need some help with implicit differentiation

  1. Dec 28, 2007 #1
    Hi, I need some help with these question, and would appreciate the help.

    1. The problem statement, all variables and given/known data
    Part 1
    Use implicit differentiation to find y'' if 2xy = y^2 Simplify and Leaev in terms of x and y.

    Part 2
    Use implicit differentiation to find y '' if xy + y^3 = 1
    Simplify your answer and leave it in terms of x and y

    2. Relevant equations

    3. The attempt at a solution
    For part 1;

    I had

    2x dy/dx + 2y = 2y dy/dx
    Took out 2's
    solved for dy/dx

    dy dx = y / [y-x]

    Part 2
    I tried the same, took d/dx to get

    dy/dx + 3y^2 dy/dx = 0 But here I get stuck because of the zero.
     
  2. jcsd
  3. Dec 28, 2007 #2
    It asked for y'' not y', let me work it but am I correct on that part?
     
  4. Dec 28, 2007 #3
    Ohh, so go through the process again? Ok, but can you help me along?
     
  5. Dec 28, 2007 #4
    Sure, so now we're at

    [tex]y'=\frac{x}{y-x}[/tex]

    Take the derivative again, quotient rule:

    [tex]y'=\frac{d}{dx}(\frac{x}{y-x})[/tex]

    so...

    [tex]y''=\frac{(y-x)\frac{d}{dx}(x)-x\frac{d}{dx}(y-x)}{(y-x)^2}[/tex]
     
    Last edited: Dec 28, 2007
  6. Dec 28, 2007 #5
    For part 2: You forgot or did the the product rule incorrectly;

    [tex]xy+y^3 =1[/tex]

    [tex]xy'+y+3y^2 y'=0[/tex]

    The zero means nothing. It's the same thing as saying x-2=0, well then x=2.

    Solving for y'

    [tex]y'=\frac{-y}{x+3y^2}[/tex]

    Now take the 2nd derivative, quotient rule:
     
    Last edited: Dec 28, 2007
  7. Dec 28, 2007 #6

    HallsofIvy

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    You don't need to solve for y' itself. If [itex]2xy= y^2[/itex], then [itex]2y+ 2xy'= 2yy'[/itex]. Now differentiate both sides of that with respect to x: [itex]2y'+ 2y'+ 2xy"= 2y'^2+ yy"[/itex] or [itex]4y'- 2y'^2= (y- 2x)y"[/itex] so [itex] y"= (4y'-2y'^2)/(y- 2x)[/itex]. Personally, I would consider that a perfectly good answer but since your problem specifically says "leave in terms of x and y", NOW use x/(y- x).
     
    Last edited: Dec 28, 2007
  8. Dec 28, 2007 #7
    Wow, thanks for clearing it up for me, I greatly appreciate the help =]
     
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