TITLE: Chain rule proof - Apostol
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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.