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Homework Help: Check my differentiation please

  1. Apr 19, 2005 #1
    [tex] y^{(3)}= 2+xy^3 [/tex]
    [tex]y^{(4)}=y^3+3xy^2y' [/tex]
    [tex]y^{(5)}=6y^2y'+6xy(y')^2+3xy^2y''[/tex]
    [tex]y^{(6)}=18y(y')^2+6y^2y''+6x(y')^3+12xyy'y''[/tex]

    It has been awhile since I have done implicit differentiation, and I am not quite sure if I have used the chain rule properly in each step. I would greatly appreciated any help you could give me on this.

    -Erik
     
  2. jcsd
  3. Apr 19, 2005 #2
    I'll go through, cant guarantee accuracy though.

    [tex] Third = xy^3 + 2[/tex]

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

    [tex] Fifth = 3y^2y' + 3((2xyy'+y^2)y' + y''(y^2x)) [/tex]

    [tex] Sixth = 6y(y')^2+y''(3y^2) + 2xyy'y''+y'''(y^2x) + 3((y''(2xyy'+y^2) + y'(y''(2xy)+2yy')+2yy') [/tex]
     
  4. Apr 19, 2005 #3
  5. Apr 19, 2005 #4
    thanks, I made a mistake somewhere in the sixth but when I went through it again I still get a different answer than yours. Does that link do implicit differentiation, I didn't see it anywhere. Anyway, it isn't that important. Again, thanks for your help.

    Dang, now I am currious...


    [tex] y^{(5)}=6y^2y'+6xy(y')^2+3xy^2y''[/tex]
    so for the individual terms we should get
    [tex] (6y^2y')'=12yy'+6y^2y''[/tex]
    [tex] (6xyy'y')'=6yy'y'+6xy'y'y'+6xyy''y'+6xyy'y'' [/tex]
    [tex] (3xy^2y'')'=3y^2y''+6xyy''y'+3xy^2y''' [/tex]
    added together, I get
    [tex] 12yy'+6y^2y''+6y(y')^2+6y(y')^3+18xyy'y''+3y^2y''+3xy^2y'''[/tex]
     
  6. Apr 19, 2005 #5
    for [tex] 6xy(y')^2 [/tex]

    I set u = 6xy, v = (y')^2

    Then:

    [tex] d(uv)/dx = v du/dx + u dv/dx [/tex]

    [tex] 6(y')^2(y+xy') + 6xy(2y'y'') [/tex]
     
  7. Apr 19, 2005 #6
    If your answer is expanded and mine is compressed they are the same for that term.
     
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