Proof of the Chain Rule

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I was wondering if someone could look over my proof of the chain rule that I wrote up while following the guidelines out of Taylor and Mann's Advanced Calculus.
I am currently self-studying Taylor and Mann's Advanced Calculus (3rd edition, specifically). I stumbled across their guidelines for a proof of the chain rule, leaving the rest of the proof up to the reader to complete.

I was wondering if someone could look over my proof, and point out any flaws I might have made.

Theorem:
Suppose g is differentiable at a point t0 of the interval α<t<β. Let x0 = g(t0), and suppose that f is differentiable at x0. Then the composite function F(x0) is differentiable at t0, and F'(t0) = f'(x0)g'(t0).

Proof:
Consider any nonzero value of Δt so small that α<t0+Δt<β and define
Δx = g(t0 + Δt) - g(t0),
Δy = f(x0 + Δx) - f(x0),
ε = Δy/Δx - f'(x0).

Notice that
F(t0 + Δt) - F(t0) / Δt
= f(x0 + Δx) - f(x0) / Δt
= Δy/Δt
= Δx(Δy/Δx - f'(x0) + f'(x0)) / Δt
= Δx(ε + f'(x0)) / Δt.

Then,
limΔt->0 [F(t0 + Δt) - F(t0) / Δt]
= limΔt->0[g(t0 + Δt) - g(t0) / Δt] * limΔt->0[ε + f'(x0)]. (1)

Since limΔt->0[Δx] = g(t0) - g(t0) = 0, limΔx->0[ε] = f'(x0) - f'(x0) = 0. So, Δx->0 and ε->0 as Δt->0. (1) becomes
F'(t0) = g'(t0)f'(x0).
 

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  • #2
PeroK
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Since limΔt->0[Δx] = g(t0) - g(t0) = 0, limΔx->0[ε] = f'(x0) - f'(x0) = 0. So, Δx->0 and ε->0 as Δt->0. (1) becomes
F'(t0) = g'(t0)f'(x0).
This bit i don't follow.

There's also a technical point that you may have ##\Delta x = 0##.

I would have expected an ##\epsilon - \delta## proof here.
 
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  • #3
pasmith
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Better is to set [tex]
\begin{split}
f(x + h) &= f(x) + hf'(x) + E_f(h) \\
g(y + k) &= g(y) + kg'(y) + E_g(k) \end{split}[/tex] where [tex]\lim_{h \to 0}\frac{E_f(h)}{h} = 0 = \lim_{k \to 0}\frac{E_g(k)}{k}[/tex] and calculate [tex]
\lim_{h \to 0} \frac{g(f(x+h)) - g(f(x))}{h}.[/tex]
 
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