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What's wrong with this proof of the chain rule

  1. May 26, 2008 #1

    quasar987

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    Can anyone see where the flaw is in the developement below, where I prove that (g o f)'(x)=g'(f(x)) instead of g'(f(x))f'(x), as it should be.

    Consider the usual hypothese under which the chain rule for real-valued function applies.

    Consider [itex] \epsilon>0[/itex]. Since g is differentiable at f(x_0), there exists [itex] \delta_1(\epsilon)>0[/itex] such that [itex]|y-f(x_0)|<\delta_1(\epsilon) [/itex] implies [itex]|g(y)-g(f(x_0))-g'(f(x_0))(y-f(x_0))|\leq\epsilon|y-f(x_0)|[/itex]

    On the other hand, f being differentiable at x_0, enjoys the Lipschitz property there. That is to say, there exists positive constants c and M such that [itex]|x-x_0|<c [/itex] implies [itex] |f(x)-f(x_0)|\leq M|x-x_0|[/itex].

    So for [itex]\delta=\min(c,\delta_1(\epsilon/M)/M)[/itex], we have that [itex]|x-x_0|<\delta [/itex] implies [itex]|f(x)-f(x_0)|\leq M|x-x_0|<M\delta\leq M\delta_1(\epsilon/M)/M= \delta_1(\epsilon/M)[/itex]

    which in turns implies that

    [tex]|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\leq(\epsilon/M) |f(x)-f(x_0)|\leq (\epsilon/M)M|x-x_0| = \epsilon |x-x_0| [/tex]

    And this is equivalent to saying that (g o f)'(x) = g'(f(x))...
     
    Last edited: May 26, 2008
  2. jcsd
  3. May 26, 2008 #2

    quasar987

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    Ok, I see it.
     
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