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Math Help - Backwards Chain Rule

  1. #1
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    Backwards Chain Rule

    I am too lazy to search for counter examples if somebody can find one that would be excellent. More excellent if he can proof it.


    Let f be a differenciable function at c and let g be a function which is defined on an open interval containing f(c) such that (g o f) is differenciable at c. Show that g is differenciable at c.
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  2. #2
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    Re:

    RE:

    The Backwards Chain Rule...
    18.3 Tricks of Integration
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Example 1:
    Let g(x) = 1/x
    Let f(x) = x + 1
    Let c = 0

    Example 2:
    Let g(x) = |x - 3|
    Let f(x) = 2x
    Let c = 3

    Argument:
    If f(x) is some function defined and differentiable on a closed interval [a,b] containing the point x = c, where f(c) = d != c and f'(c) = e, and if g(x) is some function defined on a piecewise, differentiable domain D = {x | x !=c} containing the point x = d, where g'(d) = n, then it holds that
    Although
    d/dx(g o f)(c) = g'(f(c))*f'(c) = g'(d)*e = ne <-- is defined
    d/dx(g(c)) <-- is undefined

    I'm not sure that counts as a formal proof. I've learned proofs on my own (much of it on this site), so I cannot guarantee that there are no flaws in my logic.
    Last edited by ecMathGeek; April 19th 2007 at 10:29 PM. Reason: fixed typos
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    Quote Originally Posted by qbkr21 View Post
    RE:

    The Backwards Chain Rule...
    18.3 Tricks of Integration
    My problem is more complicated then that. I cannot just use the rule without showing why it will work.
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  5. #5
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    Consider f(x)=x^2, g(x)=|x| and c=0.
    Both fog and gof are differentiable at c.
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  6. #6
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    Those examples certainly disprove it.
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