Re: ab-cd = a(b-d) + d(a-c) ? (derivatives)

Tonico <Tonicopm@xxxxxxxxx> wrote:
On Sep 18, 1:12 pm, Nando <nando.portu...@xxxxxxxxx> wrote:

II'm having troubles understanding the demonstration of the derivative
of multiplication.> The demonstration starts like this

[f(x)g(x) - f(a)g(a)] / (a - x) =  f(x) . [ ( g(x)-g(a) ) / (x- a)] + g
(x) [ f(x) - f(a) / (x-a)]

I did not really understand how to go from the first expression to the
second, so I tried to simplify it:

ab-cd = a(b-d) + d(a-c)...

Nevertheless, I still have no clue how to get from one to the other.
I don't see how to use association or distribution... Any hints?

This trick repeats over and over in several contexts. It may help you
to do the following, which is basically adding zero:

f(x)g(x) - f(a)g(a) = f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a) =

f(x)[g(x) - g(a)] + g(a)[f(x) - f(a)]

and now divide through by x-a

Yes, as I've emphasized here a few times, the product rule for derivatives
arises from the product rule for differences via difference quotients.
See [1] for a purely algebraic proof (assumes abstract algebra knowledge)
and see also [2] for further connections with congruences, limits, etc.

--Bill Dubuque

[1] sci.math, 09 Jun 2008, Linear algebra with eigenvalue AB.
http://google.com/group/sci.math/msg/3d5af9755a204f6e
http://google.com/groups?selm=y8zr6b6fd92.fsf%40nestle.csail.mit.edu

[2] sci.math, 15 Jul 2002, Congruence Modulo n property
http://google.com/group/sci.math/msg/3578c2371b2b7a46
http://google.com/groups?selm=y8z1ya4tmcy.fsf%40nestle.ai.mit.edu
.

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