6
$\begingroup$

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

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

  • 0
    @ArturoMagidin Sometimes, there just isn't enough time to go surfing. Again, thank you.2012-02-23

2 Answers 2

4

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.

  • 0
    The reason this definition is "correct" Peter is that it generalizes to higher dimensions and other meanings of derivative.2012-02-23
4

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.

  • 0
    I've sent the mail. Could you only confirm it has been delivered? Thanks.2012-02-23