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A multivariable chain rule problem

  1. Aug 21, 2009 #1
    Hello all,

    I am stuck on what seems like a rather simple problem:

    Let [tex]f:\mathbb{R}^3 \rightarrow \mathbb{R}[/tex] and [tex]g:\mathbb{R}^2\rightarrow \mathbb{R}[/tex] be differentiable. Let [tex]F:\mathbb{R}^2 \rightarrow \mathbb{R}[/tex] be defined by the equation
    [tex]F(x,y)=f(x,y,g(x,y)).[/tex]
    Find [tex]DF[/tex] in terms of the partials of [tex]f[/tex] and [tex]g[/tex].

    I would greatly appreciate any help.
     
  2. jcsd
  3. Aug 21, 2009 #2
    What work have you done on it so far? Where are you getting stuck?

    (Also, if this is homework, we have a separate forum for that).
     
  4. Aug 21, 2009 #3
    [tex] \begin{array}{ll}
    DF(x,y) &= Df(x,y,g(x,y))\\
    &= \begin{bmatrix} D_1f(x,y,g(x,y)) & D_2f(x,y,g(x,y)) & D_3f(x,y,g(x,y)) \end{bmatrix}
    \end{array}[/tex]
    I suspect that the chain rule should be used, in some way, to evaluate [tex]D_3f(x,y,g(x,y))[/tex].

    Also, this isn't homework; I just started reading Analysis on Manifolds by James Munkres a few days ago.
     
  5. Aug 21, 2009 #4
    I think I see where you are getting confused. Hint: There is no such thing as [tex]D_3F[/tex] :-o

    Since F is a function of [tex]\mathbb{R}^2[/tex], [tex]D_1F = \frac{dF}{dx}[/tex], and [tex]D_2F = \frac{dF}{dy}[/tex]. Therefore, the gradient (DF) is going to be [tex](D_1F, D_2F)[/tex].

    The notation is misleading. You probably want to write out the definition of the partial derivative, [tex]D_1 F = \frac{F(x+\epsilon, y) - F(x, y)}{\epsilon}[/tex], (and similarly for [tex]D_2 F[/tex]). Only then should you expand F out.
     
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