How do you prove the chain rule?

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Answer 1 · 2016-09-01T13:08:44

see below

i'm gonna make this up on the spot so mea culpa if it's messy but I find Taylor expansions really useful in these sorts of situations

if we start with function $f( g(x))$

and by definition we have

$(df)/(dx) = lim_(h to 0) (f(g(x+h)) - f(g(x)))/(h)$

Using Taylor, we're gonna expand the first bit as follows:

$g(x+h) = g(x) + h g'(x) + O(h^2)$

So to clarify, we have
$(df)/(dx) = lim_(h to 0) (f color(red)(( g(x) + h g'(x) + O(h^2))) - f(g(x)))/(h)$

now to simplify a little we set: $eta(x) = g'(x) + O(h)$

$(df)/(dx) = lim_(h to 0) (f ( g(x) + h eta (x) ) - f(g(x)))/(h) qquad square$

And now we're gonna expand $f ( g(x) + eta (x) )$ by the same process

$f ( g(x) + h eta (x) ) = f(g(x)) + h eta (x) f'(g(x)) + O(h^2)$

$= f(g(x)) + h ( g'(x) + O(h)) f'(g(x)) + O(h^2)$

$= f(g(x)) + h f'(g(x)) g'(x) + O(h^2)$

We can put that in $square$

$(df)/(dx) = lim_(h to 0) ( f(g(x)) + h f'(g(x)) g'(x) + O(h^2)- f(g(x)))/(h)$

$(df)/(dx) = lim_(h to 0) ( h f'(g(x)) g'(x) + O(h^2))/(h)$

$= lim_(h to 0) f'(g(x)) g'(x) + O(h)$

$= f'(g(x)) g'(x)$