Proving That d(e^x)/dx = e^x

Date: 06/30/99 at 01:20:13
From: Geoff Kissane
Subject: Calculus

I teach Year 12 calculus and we came across a question that asked to 
find the derivative of y = e^x, using the definition of the 
derivative; that is, to find

       limit  [e^(x+h) - e^x)]/h
       h -> 0

We have tried lots of things, but no luck yet.

I hope you can help.

Geoff Kissane

Date: 07/01/99 at 17:17:30
From: Doctor Fwg
Subject: Re: Calculus

Dear Geoff,

Here is one possible "Proof" that d(e^x)/dx = e^x.

First, let:

     d(e^x)/dx = lim(h->0) [e^(x+h) - e^x]/h
               = lim(h->0) e^x[e^h - 1]/h.

But:
     e^h = 1 + h/1! + h^2/2! + h^3/3! + ... + h^n/n!

So:
     [e^h - 1]/h = [h/1! + h^2/2! + h^3/3! + ... + h^n/n!]/h
                 = 1/1! + h/2! + h^2/3! + ... + h^(n-1)/n!
                 = 1.

So:
     lim(h->0) [1/1! + h/2! + h^2/3! + ... + h^(n-1)/n!] = 1.

So:
     d(e^x)/dx = lim(h-->0)e^x[e^h - 1]/h = e^x.

I hope that this helps.

With best wishes,

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/