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When considering two functions $f(x)$ and $g(x)$, it is known that

$$\left(f\circ g(x)\right)' = f'\circ g(x)\cdot g'(x)$$

So my intuitive approach is:

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {g\left( {x + \Delta x} \right)} \right) - f\left( {g\left( x \right)} \right)}}{{\Delta x}}$$

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {g\left( {x + \Delta x} \right)} \right) - f\left( {g\left( x \right)} \right)}}{{g\left( {x + \Delta x} \right) - g\left( x \right)}}\frac{{g\left( {x + \Delta x} \right) - g\left( x \right)}}{{\Delta x}}$$

Put $g\left( {x + \Delta x} \right) - g\left( x \right) = \Delta g\left( x \right)$

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{f\left( {g\left( x \right) + \Delta g\left( x \right)} \right) - f\left( {g\left( x \right)} \right)}}{{\Delta g\left( x \right)}}\frac{{g\left( {x + \Delta x} \right) - g\left( x \right)}}{{\Delta x}}$$

So I guess the problem boils down to translating how $\Delta x \to 0 \Rightarrow \Delta g\left( x \right) \to 0$ and to adress ${\Delta g\left( x \right)}$'s behaviour.

The last intuition is to recklessly write

$$g\left( {x + \Delta x} \right) - g\left( x \right) = \Delta g$$

and put

$$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta f\left( {g\left( x \right)} \right)}}{{\Delta g\left( x \right)}}\frac{{\Delta g\left( x \right)}}{{\Delta x}}$$

which is the idea behind

$$\frac{{df}}{{dx}} = \frac{{df}}{{dg}}\frac{{dg}}{{dx}}$$

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$\LaTeX$ tip: the symbol for composition is given by \circ, not by \text{ o }. –  Arturo Magidin Feb 23 '12 at 3:55
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You can try deTexify; when I draw a little circle, \circ is the top hit. And of course, there are tables of TeX symbols available all over the net. –  Arturo Magidin Feb 23 '12 at 3:57
    
@ArturoMagidin Sometimes, there just isn't enough time to go surfing. Again, thank you. –  Pedro Tamaroff Feb 23 '12 at 4:00
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2 Answers

up vote 4 down vote accepted

The chain rule is very simple, if you use the correct definition of the derivative. The derivative $f'(x)$ is a function such that $$f(x + \epsilon) = f(x) + \epsilon f'(x) + o(\epsilon).$$ If you don't know what the "little oh" notation mean, think of it as $$f(x + \epsilon) \approx f(x) + \epsilon f'(x).$$ Similarly, $$g(x + \epsilon) \approx g(x) + \epsilon g'(x).$$ Therefore, using continuity, $$f(g(x+\epsilon)) \approx f(g(x) + \epsilon g'(x)) \approx f(g(x)) + \epsilon g'(x) f'(g(x)).$$ We get the chain rule: $$(f \circ g)'(x) = f'(g(x)) g'(x).$$

The only non-trivial part is $$y \approx z \Longrightarrow f(y) \approx f(z), $$ which is a statement of continuity.

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The notes above are also contained in my video: youtube.com/watch?v=7HpqWjOcrUc proving the chain rule. –  Scott Carter Feb 23 '12 at 3:30
    
The reason this definition is "correct" Peter is that it generalizes to higher dimensions and other meanings of derivative. –  Chris Janjigian Feb 23 '12 at 3:39
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Your intuition is solid, and the fact that $\Delta x \rightarrow 0 \implies \Delta g \rightarrow 0$ follows from the continuity of $g$.

There is a subtlety though: what if $\Delta g = 0$ for arbitrarily small $\Delta x$? It is still possible to push the proof through, mostly just by thinking carefully about what this means. I do this in $\S 5.2$ of these lecture notes.

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Oh! I've found a professor (or have you found me?). Would it be too much if I borrowed some of your time in the chat? It's about this. I just want to show you my findings, not ask for help in the writing. Thanks for the .pdf - saved it and will study from it happily. –  Pedro Tamaroff Feb 23 '12 at 2:37
    
@Peter: sorry, I am not available for chat at this time. –  Pete L. Clark Feb 23 '12 at 2:42
    
I see. Can I send you an e-mail then? –  Pedro Tamaroff Feb 23 '12 at 2:48
    
@Peter: Yes, that's fine. I'm quite busy these days, so I can't promise a prompt reply... –  Pete L. Clark Feb 23 '12 at 2:54
    
I've sent the mail. Could you only confirm it has been delivered? Thanks. –  Pedro Tamaroff Feb 23 '12 at 3:41
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