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Apostol calculus I page 174-175 has the proof of chain rule.

Theorem states: Let f be the composition of two functions u and v, say $f=u \circ v$. Suppose that both derivatives $v'(x)$ and $u'(y)$ exist, where $y=v(x)$. Then derivative $f'(x)$ also exists and is given by the formula $f'(x)=u'(y).v'(x)$.

Proof: The difference quotient for f is (4.12): $\frac{f(x+h)-f(x)}{h}=\frac{u[v(x+h)]-u[v(x)]}{h}$ . Let $y=v(x)$ and let $k=v(x+h)-v(x)$. Then we have $v(x+h)=y+k$ and (4.12) becomes (4.13): $\frac{f(x+h)-f(x)}{h}=\frac{u(y+k)-u(y)}{h}$ .

If $k\neq0$,then we multiply and divide by k and obtain (4.14): $\frac{u(y+k)-u(y)}{h}\frac{k}{k}=\frac{u(y+k)-u(y)}{k}\frac{v(x+h)-v(x)}{h}$. When h goes to 0, last quotient on right becomes $v'(x)$. Also, as $h$ goes to $0$, $k$ also goes to $0$ because $k=v(x+h)-v(x)$ and $v$ is continuous at $x$. Therefore the first quotient on the right approaches $u'(y)$ as $h$ tends to zero and this proves the result. $\square$


Although the foregoing argument seems to be the most natural way to proceed, it is not completely general. Since $k=v(x+h)-v(x)$, it may happen that $k=0$ for infinitely many values of $h$ as $h$ tends to zero in which case the passage from (4.13) to (4.14) is not valid.


My doubt: I have trouble understanding the line "it may happen that $k=0$ for infinitely many values of $h$ as $h$ tends to zero" What is this line trying to convey and why is the proof incorrect?

Thanks in advance.

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  • $\begingroup$ It means that in a neighborhood of $x$ (i.e., for some $\delta>0$, $x-\delta<x<x+\delta$), $k(x)=0$ (or $v(x)=\text{constant}$). This possibility nullifies, therefore, the legitimacy of division by $k$. $\endgroup$ – Mark Viola Jul 7 '16 at 16:09
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Apostol has in mind functions like the topologist's sine curve $$t(x) = \sin \left( \frac1x \right)$$ While this function itself is not differentiable at zero, so it is not problematic for the chain rule proof, its weird cousin $$f(x) = e^{-\frac1{x^2}} \sin \left( \frac1x \right)$$ is differentiable everywhere (though it is not analytic at $x=0$).

So the question becomes, "does the chain rule apply if one of the functions is a weird function such as $f(x)$?"

Not a very practical worry, but if you present a "proof" it is always best that the proof be airtight.

Added afterward

Let $g(x) = \frac1{x+1}$. Then $$(f\circ g)(x) = e^{-(1+x)^2}\sin(1+x) \\\frac{d(f\circ g)(x))}{dx} = e^{-(1+x)^2}\cos(1+x)-2e^{-(1+x)^2}(1+x)\sin(1+x)\\ \left.\frac{d(f\circ g)(x))}{dx}\right|_{x=0} = \frac{\cos(1)-2\sin(1)}{e} \approx -0.42 \neq 0 $$ But applying the chain rule, and noting that the derivative at zero of $f(x)$ is zero,

$$ \left.\frac{df(x)}{dx}\right|_{x=0} = 0 \\ \left.\frac{dg(x)}{dx}\right|_{x=0} = -1 \frac{d(f\circ g)(x))}{dx} = 0\cdot (-1) = 0 $$

But the combination of $f$ and $g$ is not a counterexample to the chain rule, because the chain rule requires taking the derivative of $f$ at $g(x)$ and $g(0)$ is not zero.

Turns out the conditions stated in Apostol are in fact sufficient; as long as the functions are differentiable, at $g(x)$ and $x$ respectively, the chain rule works.

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  • $\begingroup$ Your example of $f$ and $g$ does not appear to be relevant to the problem under discussion. It would be better if you interchange $f,g$. $\endgroup$ – Paramanand Singh Jul 8 '16 at 5:25
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It is a common problem with proof of chain rule in calculus textbooks. It is great that your book mentions about the problem when $k=0$. Thus the given proof does not work when $v(x)= x^{2}\sin(1/x)$ and $v(0)=0$ and point of consideration is $x=0$. You should observe that $k=v(h)-v(0)=h^{2}\sin(1/h)$ vanishes at points $h=1/n\pi$ for all non zero integers $n$. This kind of behavior is what is meant by the line "$k=0$ for infinitely many values of $h$ as $h$ tends to $0$."

The proof given in your book is therefore incomplete and should handle the case when $k=0$. It might be harsh to call the proof incorrect, but rather I would term it as partial/incomplete.

However it is easy to salvage the proof when $k$ vanishes infinitely many times as $h$ tends to $0$. The thing to note is that in this case $v'(x)=0$ and we need to show that $f'(x)=0$. When $k=0$ then $f(x+h)-f(x)=0$ and if $k\neq 0$ then the ratio $(f(x+h)-f(x))/h$ can be made arbitrarily small because $v'(x)=0$. Hence $f'(x)=0$.

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  • $\begingroup$ I could understand the issue better now. Could you elaborate on "The thing to note is that in this case $v′(x)=0$ and we need to show that $f'(x)=0$"? Where in the proof are we trying to show $f'(x)=0$"? $\endgroup$ – ForumWhiner Jul 8 '16 at 13:43
  • $\begingroup$ @ForumWhiner: if $v'(x)=0$ and chain rule says $f'(x)=u'(v(x))v'(x)=0$. So we just need to see why $v'(x)=0$ in the case when $k=0$ infinitely many times and then show $f'(x)=0$ to prove chain rule. Well it is much easier to show that if $v'(x)\neq 0$ then $k=v(x+h)-v(x)\neq 0$. You should be able to prove this yourself. $\endgroup$ – Paramanand Singh Jul 8 '16 at 15:28
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Apostol's proof is common, but wrong. We need a denominator-free description of the derivative. This is provided by the following

Lemma. The function $f$ is differentiable at the interior point $a$ of its domain iff there is a function $m$ with the same domain, and continuous at $a$, satisfying $$f(x)-f(a)=m(x)\>(x-a)\ .$$ The value $m(a)$ is called the derivative of $f$ at $a$.

Denote this function by $m_{f,\,a}$ when necessary. We then have $$\eqalign{f(x)-f(a)&=u\bigl(v(x)\bigr)-u\bigl(v(a)\bigr)=m_{u,\,v(a)}\bigl(v(x)\bigr)\ \bigl(v(x)-v(a)\bigr)\cr &=m_{u,\,v(a)}\bigl(v(x)\bigr)\ m_{v,\,a}(x)\ (x-a)\ .\cr}\tag{1}$$ The factor $$g(x):=m_{u,\,v(a)}\bigl(v(x)\bigr)\ m_{v,\,a}(x)$$ in $(1)$ is continuous at $x=a$ and has the value $u'\bigl(v(a)\bigr)\>v'(a)$ there. The Lemma then implies that $(u\circ v)'(a)$ exists, and has the proposed value.

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I also had some hard time understanding this proof of Apostol. However, I do not think the answers clarify this issue. In order to understand this proof, I had to dissect it up to the epsilon-delta definition of the derivative.

Suppose you have a function $u(v(x))$ and you are looking for its derivative. However, the derivative itself is a limit:

$$f'(x) = \lim_\limits{h \rightarrow 0}\frac{u(v(x+h)) - u(v(x))}{h}$$

But the main issue with limits is their neighbourhood definition: the function does not have to exist at the central point! It is reflected in the radii definition of the limit.

Suppose the limit is at point 0 (which is the case for any derivative, since the derivative is a function of the variable $h$), then for the limit value $A$: $|f(h) - A| < \epsilon$ for any $h$ which satisfies $0 < |h| < \delta$. This is the central thing to understand. It shows that $h$ is NEVER zero. Thus, inside the limit we are justified for example to divide by $h$ in $\lim_\limits{h \rightarrow 0} \frac{mh}{h} = m$.

Now, back to the proof of Apostol. He suggests introducing the variable $k = v(x + h) - v(x)$, which we use to divide inside the limit. But the problem here is that the factor $\frac{k}{k} = k\frac{1}{k}$ can be zero for some value $h$ when $h > 0$! Think of some infinitely oscillating sinusoids with damping given by other answers here (like $x\sin\frac{1}{x}$). It can easily have $v(x + h) = v(x) \Rightarrow k = 0$ for some $h > 0$, or many such $h$'s. This multiplication by $\frac{k}{k}$ is prohibited in any limit, since this kind of expression (division by zero) is simply not defined (we define the domain of $h$ to be greater than zero, but in this domain $k = v(x + h) - v(x)$ can take zeros at arbitrary points)!

Apostol handles this issue by removing the discontinuity in the function $g(k) = \frac{u(y + k) - u(k)}{k}$, by defining $g(0) = 0$. Without this discontinuity removal $k(g(k) + u'(y))$ cannot be substituted inside the limit either! So the key step is really not even the new function $g(k)$ itself, but the fact that this function is continuous everywhere except at $k = 0$! However, its limit exists even at $k = 0$. Thus, the discontinuity can be removed by redefining $g(0) = 0$. There is no way to do this for $k = v(x + h) - v(x)$, since at these points (where $k = 0$) the function value is zero, and $k$ is in denominator. But for $g(k)$ the situation is different - there is only 1 such point. It is at $k = 0$, and $g(k)$ is in nominator! Thus it is possible to do this discontinuity removal, and limit calculation afterwards.

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