1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    Thanks!
     
  2. jcsd
  3. Jan 9, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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: Jan 9, 2013
  4. Jan 9, 2013 #3

    lurflurf

    User Avatar
    Homework Helper

    Use (derive?) Faà 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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Compute Higher Order Mixed Derivative.
Loading...