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Multivariable chain-rule proof?

  1. Mar 19, 2008 #1
    can someone link/show me a formal proof of the multivariable chain rule?
     
  2. jcsd
  3. Mar 19, 2008 #2
    There's a cute proof of this in Spivak's "Calculus on Manifolds". If you want to try it yourself, try defining the function fd(x) = f(x + h) - f(x) - df(h) and similarly for g, pretty much the numerator in the definition of the derivative, then evaluating the limit in the definition of the derivative for [itex]g\circ f[/itex], where you let the linear function in the limit be dg(f(x))df(x).
     
  4. Mar 19, 2008 #3
    what? how is this multavariable?
     
  5. Mar 20, 2008 #4

    lurflurf

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    let x be a vector x=(x1,x2,...,xn)
     
  6. Mar 20, 2008 #5
    There must be a better proof of the chain rule that that.
     
  7. Mar 20, 2008 #6
    yea i thought of that and thought the same thing; that there's gotta be a more rigorous proof.
     
  8. Mar 21, 2008 #7

    lurflurf

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    more rigorous than very rigorous
    interesting
    maybe you mean more detailed
    that I can provide

    Chain rule
    let
    f:E->F
    g:F->G
    with E (or a subset) open in F
    F (or a subset) open in G
    and f differentiable at x
    g differentiable at f(x)
    then
    (g◦f)' exist with
    (g◦f)'=[g'◦f][f']
    or in more full notation
    [g(f(x))]'=[g'(f(x)][f'(x)]
    notes on mappings
    f:E->F
    g:F->G
    f':E->L(E,F)
    g':F->L(F,G)
    (g◦f):E->G
    (g◦f)':E->L(E,G)
    g'◦f:E->(F,G)
    [g'◦f][f']:E->L(E,G)
    where L(E,F) is a space of linear mappings from E to F
    so all is as it should be
    thus derivatives are linear mappings
    Δx=dx
    Δf:=f(x+dx)-f(x)
    df:=f'(x)dx
    Δf=df+o(dx) (f differentiable)
    informal derivation
    dg(f(x))=g'(f(x))df(x)+o(df(x))=g'(f(x))f'(x)dx+o(dx)+f'(x)o(dx)
    more formal
    let
    Δf=df+|dx|rf
    rf=(Δf-df)/|dx|
    so lim rf=0
    now
    lim rf=lim rg=0 (f,g differentiable)
    Δ[g(f(x))]=g'(f(x))Δf(x)+|Δf|rg(f(x))
    =g'(f(x))f'(x)dx+|dx|r(f)+|dx||f'(x)dx/|dx|+rf|rg)
    =g'(f(x))f'(x)dx+|dx|{rf+|f'(x)dx/|dx|+rf|rg}
    we now need only
    lim {r(f)+|f'(x)dx/|dx|+rf(x)|rg(f(x))}=0
    which is clear from
    rf(x)->0
    rg(f(x))->0
    and
    |f'(x)dx/|dx|+rf|<|f'(x)|+|rf|<∞
    ie bounded near x
    where |f'(x)| is the norm induced on linear maps by the norm on vectors
     
    Last edited: Mar 21, 2008
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