Derivative of exponential function

From: Michael Stemper (mstemper_at_siemens-emis.com)
Date: 10/07/04 

Date: Thu, 7 Oct 2004 12:12:10 -0500

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? 

-- 
Michael F. Stemper
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"Writing about jazz is like dancing about architecture" - Thelonious Monk

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