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Homework Help: Compute Higher Order Mixed Derivative.

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Let f(u,v) be an infinitely differentiable function of two variables, and let g(x,y) = (x^2 + y^4, xy). If f[itex]_{v}[/itex] (5,2) = 1, f[itex]_{uu}[/itex] (5,2) = 2, f[itex]_{vv}[/itex] (5,2) = -2 and f[itex]_{uv}[/itex] (5,2) = 1, computer d^2(f o g)/dxdy at (2,1)

    2. Relevant equations

    The Chain Rule

    3. The attempt at a solution

    I set u = x^2 + y^4 and v = xy . Differentiating df/dx gives 2x(df/du) + y(df/dv) .[ df/dy gives 4y^3(df/du) + x(df/dv) but I don't think that's relevant ]. My idea from there was to use the results stated in the question to fill in these unknown derivatives.

  2. jcsd
  3. Jan 9, 2013 #2


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    Yes, [itex]\partial f/\partial x= 2x \partial f/\partial u+ y\partial f/\partial v[/itex].

    Now, differentiate that with respect to y:
    [itex]2x\partial^2 f/\partial u^2(4y^3)+ ...[/itex]
    Last edited by a moderator: Jan 9, 2013
  4. Jan 9, 2013 #3


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    Use (derive?) FaĆ  di Bruno's formula.
    Just include all possibilities
    or do
    chain rule
    product rule (twice)
    chain rule (twice)

    [tex](f \circ g)_{xy}=(f_u u_x+f_v v_x)_y \\
    = f_u u_{xy}+f_v v_{xy}+(f_u)_y u_x+(f_v)_y v_x\\
    =f_u u_{xy}+f_v v_{x y}+f_{uu} u_x u_y+f_{uv} u_x v_y+f_{vu} v_x u_y+f_{vv} v_x v_y[/tex]
    where subscripts denote differentiation and g(x,y)=(u,v)
  5. Jan 10, 2013 #4
    Ok so I'm pretty sure I got the derivative worked out. First we're trying to figure out d(f o g)/dxdy. So differentiate with respect to y first then x. Doing so gives:

    [ df/dy = 4y3 df/du + xdf/dv ]

    Differentiating with respect to x gives:

    [ 4y3d/dx(df/du) + df/dv +d/dx(df/dv) ]

    Then, differentiating d/dx(df/du) and d/dx(df/dv), and putting those back into the equation above gives:

    4y3d2f/du22x + d2f/dudvy +df/dv + d2f/dvdu2x + d2f/dv2y

    If anyone wants to link me a "how to write math properly on this forum guide" that would be swell..
  6. Jan 10, 2013 #5


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