# Help understanding Rudin's proof of the chain rule

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
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## 2 Answers

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
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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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