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Double derivate implisitt function

  1. Jul 17, 2009 #1
    We got:

    solve by x.

    y^3+x^2y^2-y*sinx=1

    derivate:

    3y^2*y' + 2xy^2 + 2yx^2*y' - (y'sinx + ycosx) = 0


    but to derivate this once again Im not sure how to proceed.

    My solution:

    [3y^2*y'+6y*y''] + [2y^2 + 4xyy'] + .....

    solutoin manual:

    6y*y'' + 2x2yy' + [2x^2*1*y'+2x^2*y*y'']



    I've been searching for websites where I can read about this, but no luck.

    What I think is done in the solution manual is that when x is x^1 it cannot be further derivated, because we kinda "loose" information?

    I need some guidance or explination on how to handle equations that got:

    y*y'

    x*y^2

    x*y*y'


    Thanks in advance!
     
  2. jcsd
  3. Jul 17, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Okay, good!


    [quiote]but to derivate this once again Im not sure how to proceed.

    My solution:

    [3y^2*y'+6y*y''] + [2y^2 + 4xyy'] + .....[/quote]
    You have the y' and y" reversed in the first braces
    It is (3y^2)(y')'+ (3y^2)'(y')= (3y^2)y"+ 6yy'.

    For the second (2xy^2)'= (2x)'(y^2)+ (2x)(y^2)'= 2y^2+ (2x)(2yy')= 2y^2+ 4xyy'.

    For the third term (2yx^2y')'= (2y)'(x^2)(y')+ (2y)(x)'(y')+ (2y)(x)(y')'= 2x^2y'+ 2yy'+ 2yxy".

    For the fourth term (-y'sin x)'= (-y')'(sin x)+ (-y)(sin x)'= -y"sin x- y cos x.

    For the fourth term (-ycos x)'= (-y)'(cos x)+ (-y)(cos x)'= -y' cos x+ y sin x.

    Combine those.

     
  4. Jul 18, 2009 #3
    Thanks! That made a lot more sense than the teachers solution. I get it 100% now =)
     
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