On Mon, 16 Nov 2009 15:11:13 +0100, Invisible <voi### [at] devnull> wrote:
>
> I still don't entirely understand what you're saying

(Excuse the mixing of Lagrange notation and ASCII-fied Leibniz notation)

First look at y = f(x). Then f'(x) = dy/dx.

Now try y = f( g(x) )

Let u = g(x). Then y = f( u ).

In f'(x) = dy/dx, we can trivially introduce 'u' thus:

  f'(x) = dy/dx * (du/du) = (dy/du) * (du/dx)

dy/du is obviously the derivative of 'y' with respect to 'u', i.e. f'(u).

Similarly, du/dx is the derivative of 'u' with respect to 'x', i.e. g'(x).

  f'(x) = (dy/du) * (du/dx) = f'(u) * g'(x)

Substitute u = g(x) to get:

  f'(x) = f'( g(x) ) * g'(x)



-- 
FE

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