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So I was looking through various proofs of the chain rule...and I came across this paper. The first proof given is complete and quite well-explained. But another simplistic proof is given in the end...which is mentioned as "technically incorrect". Can anyone tell me why?

Here is the incorrect proof in question:

$$\begin{aligned} (f \circ g)'(x) &= \lim_{h \to 0}\frac{f(g(x+h)) - f(g(x))}{h} \\ \implies (f \circ g)'(x) \cdot \left(\frac{1}{g'(x)}\right) &= \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{h}\right)\cdot\left(\frac{h}{g(x+h)-g(x)}\right)\\ &= \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}\right) \\ &= f'(g(x)) \\ \end{aligned}$$

SirXYZ
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    The manipulation at the end of the simplistic proof is wrong whenever $g(x+h) = g(x)$. It also assumes implicitly that $g'(x) \neq 0$. –  Nov 11 '16 at 17:51
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    One more problem: The second line is using the fact that the product of two limits (the derivatives on the LHS) equals the limit of the products (the RHS). However, you can't conclude this unless you already know that both limits (i.e. both derivatives on the LHS) exist. –  Nov 11 '16 at 18:01
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    @Bungo how about making this an answer? This is the more relevant problem about this proof. That about $g'(x) = 0$ could easily just be covered as a separate case. – leftaroundabout Nov 11 '16 at 19:02
  • @leftaroundabout I expanded my comment into an answer. –  Nov 11 '16 at 19:12
  • By "techincially incorrect" I assume they simply mean "incorrect". – zhw. Nov 11 '16 at 19:53

2 Answers2

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If $g'(x) = 0$, the proof fails, but as mentioned in the comments, this can be handled as a separate case.

A more serious issue is the calculation $$(f \circ g)'(x) \cdot \left(\frac{1}{g'(x)}\right) = \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{h}\right)\cdot\left(\frac{h}{g(x+h)-g(x)}\right)$$ If we write $$a(h) = \frac{f(g(x+h))-f(g(x))}{h}$$ and $$b(h) = \frac{h}{g(x+h)-g(x)}$$ then the above calculation is the assertion that $$\lim_{h \to 0}\ a(h) \cdot \lim_{h \to 0}\ b(h) = \lim_{h \to 0}\ a(h)b(h),$$ in other words, that the limit of a product is the product of the limits. This is true, provided that both limits on the left-hand side exist. But the existence of $\displaystyle \lim_{h \to 0}\ a(h)$ is exactly what we are trying to prove, so the argument is circular.

Edit to add:

Another issue worth noting is the final assertion, namely $$f'(g(x)) = \lim_{h \to 0}\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}$$ In fact, the definition of the derivative of $f$, evaluated at $g(x)$, gives us $$f'(g(x)) = \lim_{k \to 0}\frac{f(g(x)+k) - f(g(x))}{k}$$ The only reason we are able to conclude that these two expressions are equal (even after handling the $g(x+h) = g(x)$ case properly) is because $g$ is continuous at $x$. This of course follows from the differentiability of $g$ at $x$, but a careful proof would point this out.

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    As long as $\lim b(h) \neq 0$ I can write $\lim a(h) = \lim a(h)b(h) \cdot \lim b(h)^{-1}$ to prove the existence (provided both $\lim b(h)$ and $\lim a(h)b(h)$ exist, which is the case here). So this is actually a less serious issue than the forgotten special case, imho. Am I mistaken somewhere? – Hermann Döppes Nov 11 '16 at 23:34
  • @HermannDöppes I guess which flaw is more serious is debatable. Incorrect application of the algebra of limits is a common error, and can lead to incorrect results. In this case, the argument can of course be modified to make it correct (the theorem is, after all, true), but the question is why this proof is technically incorrect, and I think it's worth calling attention to this issue. –  Nov 11 '16 at 23:46
  • I agree on almost all points, but want to make two on my own: 1. Not every argument leading to a correct result (i.e. a true theorem) can necessarily be fixed. I don't know of a counter example right know but I'm pretty sure such a thing would exist. – Hermann Döppes Nov 12 '16 at 00:23
  • While it is perfectly debatable, I just want to give my two reasons for considering one more serious than the other: One is a trick I have seen multiple times already, which is applicable to many situations and should be part of the curriculum or exercise sheets of any math student. The other is something I could not solve in under 5 seconds. YMMV, of course. This is all highly subjective.
    • – Hermann Döppes Nov 12 '16 at 00:28