Re: Derivative of exponential function
From: Virgil (ITSnetNOTcom#virgil_at_COMCAST.com)
Date: 10/07/04
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Date: Thu, 07 Oct 2004 12:29:49 -0600
In article <200410071712.i97HCAL62738@mickey.empros.com>,
mstemper@siemens-emis.com (Michael Stemper) wrote:
> I'm in the process of home-schooling my son in calculus. We're about to
> finish up on limits, so that we can start differentiation. I've relearned
> enough that I've been able to derive, from the definition of derivative,
> the derivatives of c, x, cx, f(x)+g(x), f(x)*g(x), f(g(x)), sin(x),
> cos(x). From those, I can get the derivatives of f(x)-g(x), f(x)/g(x),
> (f(x))^n, (f(x))^(-n), tan(x). A pretty solid beginning. But there's
> one big hole: I can't prove that the derivative of e^x is e^x.
>
> I can see that if Lim(h->0) ((e^h-1)/h) == 1, then I'm set. Numerically,
> I can crank out values for ((e^h-1)/h) with h getting very small, and
> see that they get really close to 1. But, that's not proof.
>
> I was going to base a proof on the Taylor series for e^x, but that
> depends upon already knowing the nth derivatives of e^x, so that was
> out. Then I was going to try using my knowledge of the behavior of
> e^x, but upon examination, that was all based on knowing things about
> its derivative.
>
> Is there some simple (or subtle) trick that I'm overlooking? Is the
> proof of this limit actually incredibly hard?
>
> I can't look in the book, because it's with him (he lives with his
> mother). I've even tried typing "((e^h-1)/h)" into Google, but that
> just turned up a bunch of PDF files. Any help, or am I going to need
> to do some serious hand-waving?
How did you define e^x in the first place?
The two commonest definitions, at least as far as I recollect, are
(1) by the power series and (2) as the inverse function to the natural
logarithm function (in turn defined as log(x) = integral[t=1..x,1/t],
for x > 0).
Case (1) you say you know how to do.
In case 2, apply the chain rule to x = log(e^x))
If you have some other definition, if you post your definition, maybe
someone will be able to help.
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