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

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> **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.