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Calculus chain rule

  1. Aug 14, 2012 #1
    Hi, I have been doing research in my spare time this summer on calculus proofs. I am working on a mathematics degree and I am working to understand calculus inside and out. It has been going really well but I have sort of hit a bump with the calc 1 chain rule. Here is my attempt:

    lim h -> 0 [f(g(x+h))] - f(g(x))]/h is what I am aiming to solve. We want to work from the inside out, so lets start with g(x).

    The derivative of g(x) is lim h -> 0 [g(x+h) - g(x)]/h. Since this is the derivative of g(x) I can rewrite this as [g(x+h) - g(x)]/h = g'(x). Now I solve for g(x+h). Multiply both sides by h. [g(x+h) - g(x)] = h[g'(x)]. Add g(x) to both sides. g(x+h) = h[g'(x)] + g(x). Now I know what g(x+h) equals.

    Likewise I have to know what f(x) is too. I will be using y and k for f(x) since I used x and h for g(x). The derivative of f(y) in this case is lim h -> 0 [f(y+k) - f(y)]/k. Using the same idea as above to solve for f(y+k), I get f(y+k) = k[f'(y)] + f(y).

    So f([h[g'(x)] + g(x)] - f(g(x))/h. Now I am unsure where to go from here.
     
  2. jcsd
  3. Aug 14, 2012 #2

    LCKurtz

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    No, you can't do that. Those aren't equal. What you could write is$$
    g'(x) = \frac{g(x+h)- g(x)}{h} +\left( g'(x) - \frac{g(x+h)- g(x)}{h}\right)$$The expression in parentheses approaches 0 as ##h\rightarrow 0## so for convenience, just call the expression in parentheses ##\epsilon(h)## so you can write$$
    g'(x) = \frac{g(x+h)- g(x)}{h} +\epsilon(h)$$where ##lim_{h\rightarrow 0}\epsilon(h)=0##.

    Look at this link: http://en.wikipedia.org/wiki/Chain_rule and scroll down to the second proof of the chain rule to see how this idea is used to make a formal proof.
     
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