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Implicit Differentiation

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider:
    [tex]x^3+y^3+2xy=4[/tex], y=1 when x=1
    a.) Find the equation of the tangent line to the curve when x=1.
    b.) Find y'' at x=1.
    c.) Is the graph of y=f(x) concave up or concave down near x=1?

    2. Relevant equations
    Any derivative rules...

    3. The attempt at a solution
    For Part a:

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

    After substituting x=1 and y=1:

    [tex]y'=-1[/tex]

    For y=mx+B, B=2

    [tex]y=-x+2[/tex]

    For Part b:

    [tex]y''=\frac{vu'-uv'}{v^2}[/tex]

    [tex]u=-2y-3x^2[/tex]
    [tex]u'=-2y'-6x[/tex]
    [tex]v=3y^2+2x[/tex]
    [tex]v'=6yy'+2[/tex]

    [tex]y''=\frac{-(3y^2+2x)(2y'+6x)+(6yy'+2)(2y+3x^2)}{(3y^2+2x)^2}[/tex]

    After substituting x=1, y=1, and y'=-1:

    [tex]y''=-1.6[/tex]

    For Part c:

    Would it be concave down because I just tested y'' with x=1?

    I'm mostly unsure about my steps in part b.) and c.), and I want to make sure I'm doing it correctly. Thanks for your help.
     
  2. jcsd
  3. Oct 24, 2009 #2
    Sorry for this unnecessary post, but I don't have much time and need to be sure this is correct.
     
  4. Oct 25, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is right. But I wouldn't bother to solve for the general expression for y'.
    By implicit differentiation you got 3x2+ 3y2y'+ 2y+ 2xy'= 0. Setting x= y= 1 in that, 3+ 3y+ 2+ 3y'= 5+ 5y'= 0 which gives y'= -1.

    Yes, that is correct.

    Again, I would work from 3x2+ 3y2y'+ 2y+ 2xy'= 0. Differentiating both sides of that, with respect to x, 6x+ 6y(y')2+ 3yy"+ 2y'+ 2y'+ 2xy"= 0.

    Again, setting x= y= 1 and y'= -1,
    6(1)+ 6(1)(-1)2+ 3(1)y"+ 4(-1)+ 2(1)y"= 0
    6+ 6+ 3y"- 4+ 2y"= 0 gives 5y"= -8, y"= -8/5.

    Yes, -8/5= -1.6.

    Yes, y"< 0 so the graph is concave down.

     
  5. Oct 25, 2009 #4
    Thank you so much, that makes a lot more sense to do it that way. I was wondering though, for your second derivative, is 3yy" supposed to be 3y2y" or is there something I'm forgetting? Thanks again.
     
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