Is My Proof of the Chain Rule Correct?

In summary,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 defineΔx = g(t0 + Δt) - g(t0),Δy = f(x0 + Δx) - f(
  • #1
Expiring
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TL;DR Summary
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
Expiring said:
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
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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Related to Is My Proof of the Chain Rule Correct?

1. How do I know if my proof of the chain rule is correct?

One way to check the correctness of your proof of the chain rule is by comparing it to established proofs and seeing if they follow the same logic and steps.

2. What are the common mistakes to avoid when proving the chain rule?

Some common mistakes to avoid when proving the chain rule include not considering all the variables, not properly applying the chain rule formula, and not showing all the steps of the proof.

3. Can I use different approaches to prove the chain rule?

Yes, there are different approaches to proving the chain rule, such as using the limit definition of the derivative or using the power rule. As long as the logic and steps are correct, the approach can vary.

4. How can I improve my understanding of the chain rule?

To improve your understanding of the chain rule, it is important to practice solving problems that involve the chain rule and to review different proofs and approaches. You can also seek help from a tutor or attend a workshop on the topic.

5. Is it necessary to memorize the proof of the chain rule?

While it is not necessary to memorize the proof of the chain rule, it is important to understand the logic and steps of the proof. This will help you apply the chain rule correctly and efficiently in problem-solving.

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