Understanding Proof of Chain Rule Reading through this proof of the chain rule, I'm fine up until the following step:
$f[u(x+h)]-f[u(x)] = f[(ux)+h((v(h)+u'(x))]-f[u(x)]$
$= f[u(x)]+h(v(h)+u'(x))(w(k)+f'[u(x)])-f[u(x)]$
$= h(v(h)+u'(x))(w(k)+f'[u(x)])$
How does the proof go from the first line to the second? 
 A: This is just substitution.
(2) is the statement

$f(z+k)=f(z)+k(w(k)+f'(z))$;

think of this as saying that $w$ measures the error in approximating the change in $f$ (over a large region) by $f'$ (which is a purely local thing). The initial value is $z$; remember that$f(z)+kf'(z)$ is a good approximation to $f(z+k)$ if $z+k$ is near $z$, that is, if $k$ is near $0$. Meanwhile, the bigger $k$ gets, the larger the region we're looking at (going from $z$ to $z+k$), and the larger potential error. Intuitively, you often want to think of $w$ as increasing - the further away from $z$ you get, the worse this approximation becomes - although this won't always be the case: a stopped clock is right twice a day, and $\sin(10+17129384798324\pi)=\sin(10)$.
Now we're interested in a specific initial value, and a specific change; so we're going to replace $z$ and $k$ with these expressions. Specifically, we're interested in $k=h(v(h)+u'(x))$, and $z=u(x)$. We get $$f(u(x)+h(v(h)+u'(x)))=f(z+k)=f(z)+k(w(k)+f'(z))$$ $$=f(u(x))+(h(v(h)+u'(x)))(w(k)+f'(u(x))).$$Now subtract a $f(u(x))$ from both sides of $$f(u(x)+h(v(h)+u'(x)))=f(u(x))+(h(v(h)+u'(x)))(w(k)+f'(u(x)))$$ and you get from the first line to the second.
A: I think that this is a really ugly proof.
I like this one much better.
$\frac {d}{dx}f(u(x)) =$$ \lim_\limits{a\to x} \frac {f(u(x)) - f(u(a))}{x-a}\\
\lim_\limits{a\to x} \frac {f(u(x)) - f(u(a))}{u(x)-u(a)}\frac {u(x) -u(a)}{x-a}\\
\lim_\limits{a\to x} \frac {f(u(x)) - f(u(a))}{u(x)-u(a)}\lim_\limits{a\to x}\frac {u(x) -u(a)}{x-a}$
The limit on the right equals $\frac {du}{dx}$ by definition.
And since $u$ is continuous we can say:
$\lim_\limits{u(a)\to u(x)} \frac {f(u(x)) - f(u(a))}{u(x)-u(a)}\frac {du}{dx} = \frac {df}{du}\frac{du}{dx}$
