1

This proof of the one dimensional chain rule was pointed out as inaccurate (if not utterly wrong) by our calculus teacher a couple years ago: denoting the composition as $f(g(x))=(f\circ g)(x)=f(y)$ where $y\doteq g(x)$, $$\frac{\mathrm{d}f(g(x))}{\mathrm{d}x}\Big|_{x=x_0}= \lim_{x\to x_0}\frac{f(g(x))-f(g(x_0))}{x-x_0}= \lim_{x\to x_0}\frac{f(g(x))-f(g(x_0))}{g(x)-g(x_0)}\cdot\frac{g(x)-g(x_0)}{x-x_0}= \frac{\mathrm{d}f(y)}{\mathrm{d}y}\Big|_{y=g(x_0)} \frac{\mathrm{d}g(x)}{\mathrm{d}x}\Big|_{x=x_0}. $$ The reason of this remark was, of course, the possibility of dividing by zero in the second passage, given for example by a constant $g(x)$.

Is there really no way of making this statement rigorous, e.g. by sorting the possible pathological cases and so on?

Brightsun
  • 7,063

2 Answers2

1

The problem is that you do not know if $g(x_0) - g(x)$ is different from $0$ for all values of $x\neq x_0$.

Another problem is that you have no proof that

$$\lim_{x\to x_0}\frac{f(g(x))-f(g(x_0))}{g(x)-g(x_0)} = \lim_{y\to y_0}\frac{f(y)-f(y_0)}{y-y_0}$$

This equality is true for the functions you have, however, it is not trivial. Therefore, unless you prove it or justify it by some other result, you cannot simpliy claim to know it.

5xum
  • 126,794
  • 6
  • 138
  • 213
  • 1
    Does the second statement really need to be proved? I mean we are just calling $g(x):=y$ – Guy Mar 22 '14 at 13:05
  • One quick rule of thumb: if you think a statement is completely trivial and does not need proving, try and prove it in one line. If you fail, it is not trivial. – 5xum Mar 22 '14 at 13:42
  • That is what is throwing me off here, $y=g(x)\implies f(g(x))=f(y)$ and $y_0=g(x_0)\implies f(y_0)=f(g(x_0))$. Done right? – Guy Mar 22 '14 at 13:47
  • Not really, you proved nothing about limits here. Sorry, but I would not accept your argument as proof. For example, if $g(x) = 1$ for $x\neq 0$ and $g(x_0) = 0$, the equality I wrote does not hold. – 5xum Mar 22 '14 at 14:32
  • That makes $g$ discontinuous right? Wasn't continuity an implicit assumption? What are your views on this? – Guy Mar 22 '14 at 14:37
  • Of course, you have to use continuity. But this means that whatever the proof is, it is no longer "trivial", so the statement is not "obvious", don't you think so? I am not saying that the statement is not true, just that it takes some (nonzero) amount of work to prove. – 5xum Mar 22 '14 at 15:20
  • Ah. I see your point. Not "trivial" or "obvious" but you should see this nevertheless. I understand that you do, you are making the point for someone who might not. Am I right? – Guy Mar 22 '14 at 15:51
  • Yes, sorry if I did not make myself clear about that. Hopefully, this discussion will still be of use to someone. I will also try to elaborate my answer a bit more. – 5xum Mar 22 '14 at 16:09
  • The discussion will most definitely be of use. Cheers! – Guy Mar 22 '14 at 16:12
0

It turns out the answer is "yes" and the trick is shown on the very wikipedia page on the subject.

Let: $$ Q(y)= \left\{ \begin{array}{rl} \frac{f(y)-f(g(x_0))}{y-g(x_0)} & \text{if } y\not= g(x_0),\\ \frac{\mathrm{d}f}{\mathrm{d}y}\Big|_{y=g(x_0)} & \text{if } y= g(x_0). \end{array} \right. $$ Now the difference quotient trivially satisfies the following identity: $$ \frac{f(g(x))-f(g(x_0))}{x-x_0}=Q(y)\cdot \frac{g(x)-g(x_0)}{x-x_0}. $$ To conclude: $$ \lim_{x\to x_0}\frac{f(g(x))-f(g(x_0))}{x-x_0} = \lim_{x\to x_0} Q(y)\cdot \frac{g(x)-g(x_0)}{x-x_0} =_{\text{if they exist}}\lim_{y\to g(x_0)}Q(y) \lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0} = \frac{\mathrm{d}f(g(x_0))}{\mathrm{d}y}\frac{\mathrm{d}g(x_0)}{\mathrm{d}x}. $$

Brightsun
  • 7,063