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Proof of chain rule

  1. Jun 21, 2009 #1
    Hello everyone,

    I was looking at the proof of chain rule as posted here:

    http://web.mit.edu/wwmath/calculus/differentiation/chain-proof.html"

    I am having trouble understanding why delta(u) tends to 0 as delta(x) tends to 0. Can someone point out to me why that is so?

    Many thanks,

    Luca
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jun 21, 2009 #2
    Because the limit [tex]\lim_{\Delta x \to 0} \frac {\Delta u}{\Delta x}[/tex] must exist by hypothesis, and the only way that can happen is if [tex]\Delta u[/tex] decreases as [tex]\Delta x[/tex] decreases. The limit doesn't need to be 1 because the rate that the denominator and numerator decrease can differ, but it does need to be finite. Think of what would happen if [tex]\Delta u[/tex] approached a nonzero number or diverged to infinity; the limit would also diverge or not exist.
     
    Last edited: Jun 21, 2009
  4. Jun 21, 2009 #3
    u is continuous (at at least one point, the point where it is differentiable), which means
    [tex]\lim_{x\rightarrow a} u(x) = u(a)[/tex]
    for all constants 'a' at which u is continuous which is equivalent to your statement (u(x) approaches u(a) as x approaches a).
    Expanding the deltas in your limit we have the statement
    [tex]\lim_{x\rightarrow x_0} \frac{u(x) - u(x_0)}{x - x_0}[/tex]
    where a = x0.
     
    Last edited by a moderator: Apr 24, 2017
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