Proving That d(e^x)/dx = e^xDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/