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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?

Dunka
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2 Answers2

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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.

Noah Schweber
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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:

$\lim_\limits{u(a)\to u(x)} \frac {f(u(x)) - f(u(a))}{u(x)-u(a)}\frac {du}{dx} = \frac {df}{du}\frac{du}{dx}$

Doug M
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    This is, of course, not a valid proof. $u(x)-u(a)$ could be zero infinitely often as $a\to x$. – Ted Shifrin Apr 06 '17 at 23:14
  • @TedShifrin Can you help me out with an example of $u(x)$ that breaks this? $u(x) - u(a)$ goes through 0 infintely many times, while $\frac {du}{dx}$ continues to be defined and $\frac {du}{dx}$ is non-zero. – Doug M Apr 06 '17 at 23:29
  • The standard example of such a differentiable function is $$g(x)=\begin{cases} x^2\sin(1/x), & x\ne 0 \ 0, & x=0\end{cases}.$$ The tangent line as best linear approximation is really the best proof, and it generalizes to multivariable calculus, as well. – Ted Shifrin Apr 06 '17 at 23:39
  • @TedShifrin could you link me to this best proof? – Dunka Apr 06 '17 at 23:57
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    See, for example, the not-accepted answer at this question. – Ted Shifrin Apr 07 '17 at 00:16