1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook