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Second derivatives using implicit differentiation

  1. Oct 17, 2010 #1
    Hi there. Well, I wanted to know how to find the second derivatives of a function using implicit differentiation. Is it possible? I think it is. I think I must use the chain rule somehow, but I don't know how... I'm in multivariable calculus since the function I'm gonna use could be seen as a function of only one variable.

    An ellipse: [tex]F(x,y)=4x^2+y^2-25=0[/tex]

    So we have the partial derivatives:
    [tex]F_x=8x[/tex], [tex]F_y=2y[/tex]
    [tex]F_{xx}=8[/tex], [tex]F_{yy}=2[/tex]

    So then, using implicit differentiation:

    [tex]\frac{{\partial x}}{{\partial y}}=-\displaystyle\frac{\frac{{\partial F}}{{\partial y}}}{\frac{{\partial F}}{{\partial x}}}=\displaystyle\frac{-y}{4x}[/tex]

    But now if I wanna find [tex]\frac{{\partial^2 x}}{{\partial y^2}}[/tex] how should I proceed?
     
  2. jcsd
  3. Oct 17, 2010 #2

    Mark44

    Staff: Mentor

    I believe you're going about this in the wrong way, since partial derivatives are not required.

    If 4x2 + y2 + 25 = 0,
    then 8x + 2y*y' = 0 ==> y' = -4x/y

    To get y'', simply take the derivative with respect to x of -4x/y, using the quotient rule and the chain rule.
     
  4. Oct 18, 2010 #3

    HallsofIvy

    User Avatar
    Science Advisor

    Or, much the same way, but, I think, a bit more in keeping with the spirit of "implicit differentiation", from [itex]4x^2+ y^2+ 25= 0[/itex], [itex]8x+ 2y y'= 0[/itex] and differentiating again, [itex]8+ 2y'(y')+ 2yy"=[/itex][itex] 8+ 2y'^2+ 2yy"= 0[/itex] and solve that for y".
     
  5. Oct 18, 2010 #4
    Thanks!
     
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