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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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 $f'(x)=0$.

  • $\begingroup$ 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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.