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The first step of the proof of the chain rule in Rudin's Principles of Mathematical Analysis (Theorem 5.5, page 105) is as follows

Theorem. Suppose $f$ is continuous on $[a,b]$, $f'(x)$ exists at some point $x\in[a,b]$, $g$ is defined on an interval $I$ which contains the range of $f$, and $g$ is differentiable at the point $f(x)$. If $$h(t)=g(f(t))\quad (a\leq t\leq b)$$then $h$ is differentiable at $x$, and $$h'(x)=g'(f(x))f'(x)$$ Proof. Let $y=f(x)$. By the definition of the derivative, we have $$f(t)-f(x)=(t-x)[f'(x)+u(t)]$$ $$ g(s)-g(y)=(s-y)[g'(y)+v(s)]$$ where $t\in[a,b]$, $s\in I$, and $u(t)\rightarrow 0$ as $t \rightarrow x$, $v(s) \rightarrow 0$ as $s\rightarrow y$.

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I think I can follow the rest from here, but I don't understand this manipulation. The definition of the derivative gives $$f'(x)=\lim_{t\rightarrow x} \frac{f(t)-f(x)}{t-x}$$ I can sort of see what's going on—it's a little like we're multiplying both sides of the equation by $t-x$ and $u(t)$ is there to make doing that make sense but I can't figure out how.

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Wow wait I might have figured it out, hah. For fixed t, $\frac{f(t)-f(x)}{t-x}=f'(x)+\text{something}$ where (something) goes to 0 as $x\rightarrow t$. Is that right? –  crf Dec 2 '12 at 8:28
    
Am I the only one who feels $\varepsilon_1(t)$ and $\varepsilon_2(s)$ would've been more reasonable and implicative here? –  000 Dec 2 '12 at 8:51
    
@Limitless it at least would have been more suggestive –  crf Dec 2 '12 at 8:53

2 Answers 2

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What Rudin really means is this: define $$u(t)=\cases{ \frac{f(t)-f(x)}{t-x}-f'(x) & if $t \ne x$, \\ 0 & if $t = x$. }$$ for $t$ near $x$. You can see that $u(t) \to 0$ as $t \to x$ by the definition of the derivative of $f$ at $x$. Clearly, $$f(t)-f(x)=(t-x)[f'(x)+u(t)]$$ as well.

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You mean for $t$ near $x$ right? –  Nameless Dec 2 '12 at 10:38
    
@Nameless: Thanks. –  wj32 Dec 2 '12 at 10:43

As you say the definition of the derivative is

$$f'(x) = \lim_{t\rightarrow x} \frac{ f(t) - f(x)}{t-x}$$

So when $t$ is close to $x$ we know that $f'(x)$ is close to $\frac{f(t) - f(x)}{t-x}$. In fact we can judge how close it is, and say that $$f'(x) = \frac{f(t) - f(x)}{t-x} - u(t)$$ where $u(t)$ is a function that tells us the error in this approximation, or how close this is to the actual derivative.

So think of $u$ and $v$ in that proof as judging the error in approximation of $f'(x)$ here.

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yeahh I just figured that out immediately after I wrote all this stuff out. –  crf Dec 2 '12 at 8:28
    
@crf thats good, I'm glad I could confirm your thinking then. –  Deven Ware Dec 2 '12 at 8:31

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