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Homework Help: Total differential for finding higer row derivatives

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Well, let's take F: [tex] x^2 y^3=0 [/tex].
    Now, let's say thay y=y(x), y being an implicit function of x.
    I want to find 2nd row derivative [tex] \frac{d^2y}{dx^2} [/tex]
    using differential operator.


    2. Relevant equations
    not apply


    3. The attempt at a solution
    Using D for the first time:
    [tex]
    2xy^3dx+3x^2y^2dy=0
    [/tex]
    Now I can find dy/dx:
    [tex]
    \frac{dy}{dx}=-\frac{2xy}{3x^2}
    [/tex]

    pretty simple, huh?

    Now, using D for the 2nd time:

    [tex]
    2y^3dx^2+2xy^3d^2x+12xy^2dxdy+6x^2ydy^2+3x^2y^2d^2y=0
    [/tex]

    Now, the question is: how to find the value of [tex] \frac{d^2y}{dx^2} [/tex] from the equation above. I know how to do it in another way, but I struggle to use that one.

    Thanks in advance.
     
  2. jcsd
  3. Apr 22, 2010 #2

    Mark44

    Staff: Mentor

    Why don't you start with this:
    [tex]
    \frac{dy}{dx}=-\frac{2xy}{3x^2}
    [/tex]
    and take the derivative implicitly with respect to x of both sides? Is this the other technique that you're struggling with?
     
  4. Apr 22, 2010 #3
    I must have been misunderstood, apparently the lack of technical vocabulary can cause problems ;).

    I'm familiar with the method you suggested, it's pretty simple and obvious. I was just wondering how to find the value of [tex]\frac{d^2y}{dx^2} [\tex] from the equation I got after differentiating the equation above totally for the 2nd time. I mean, it has to be possible somehow :)
     
  5. Apr 22, 2010 #4

    Mark44

    Staff: Mentor

    I don't remember ever seeing anyone do what you're trying to do.
     
  6. Apr 22, 2010 #5
    Now, why I'm asking this. Of course it's as pointless as it may seem, but it seems to be quite a reasonable questions if a system of equations is to be considered.
    Like...

    u+v=x+y
    x*sin(u)=y*sin(v)

    where u=u(x,y), v=v(x,y). Now it's not that easy to find [tex] \frac{d^2u}{dx^2} or \frac{d^2v}{dy^2} [/tex] or whatever :). I mean, using total differentiation, one could easily handle this as a system of linear equations. At least I think so ;)
     
  7. Apr 22, 2010 #6
    Figured out the answer myself, that was a pretty tough puzzle, though :).
    Lets's take our equation, keeping in mind, that y=y(x):
    [tex]
    2y^3dx^2+2xy^3d^2x+12xy^2dxdy+6x^2ydy^2+3x^2y^2d^2y=0
    [/tex]
    Now, as x is a linear function of x, the 2nd differential of x disappears, hence:
    [tex]
    2xy^3d^2x=0
    [/tex]
    now:
    [tex]
    dy=\frac{dy}{dx} dx
    [/tex]
    With that we can substitute all dy's (dy^2=dy*dy=(dy)^2 -obviously):
    [tex]
    2y^3dx^2+12xy^2 \frac{dy}{dx} dx^2 + 6x^2y (\frac{dy}{dx})^2 dx^2 + 3x^2y^2d^2y=0
    [/tex]
    Then simplify and it's now easy to find
    [tex]
    \frac{d^2y}{dx^2}
    [/tex]

    Anyway, thanks for help, Mark44 :)
     
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