Derivative of e^x by Limit DefinitionDate: 05/21/2002 at 22:40:18 From: Jeff King Subject: derivative of e^x by limit definition Dear Dr. Math, I've tried (as has my entire calculus class) to prove that the derivative of e^x is e^x by the limit definition of the derivative (without using the Taylor expansion for e^x) and we cannot seem to get past one last step. In setting up the form of the limit definition of the derivative we've made e(x+delta x) into e^x*e^(delta x) for convenience and left the -e^x in the numerator of the limit definition. Then by factoring we determined that the limit as delta x goes to 0 is (e^x(e^(delta x)-1))/delta x. The e^x alone in the numerator would seem to be our best bet, but we can't make the rest of the function go to 1 as some people have guessed it should. How would one prove the limit definition or just that (e^(delta x)-1)/(delta x) = 1 as x tends to infinity? Thanks for your assistance! Date: 05/29/2002 at 22:41:49 From: Doctor Fwg Subject: Re: derivative of e^x by limit definition Dear Jeff, Here is a possible solution that does not involve the use of a Taylor expansion. I hope the notation I have used here will be easy enough to follow. Definition of e is: 1) e = Limit (1 + 1/n)^n n->Inf However, if n = 1/h, then 2) e = Limit (1 + h)^(1/h) h->0 If f(x) = e^x, the definition of f'(x) is: 3) f'(x) = Limit [f(x + h) - f(x)]/h h->0 But: 4) f(x + h) = e^(x+h) = (e^x)(e^h) So: 5) f'(x) = Limit [(e^x)(e^h) - e^x]/h h->0 = Limit [(e^x)((e^h) - 1)]/h h->0 But, raising both sides of equation (2) to the power of h yields: 6) e^h = Limit [(1 + h)^(1/h)]^h h->0 = Limit [(1+ h)] h->0 Placing this value of e^h into equation (5) yields: 7) f'(x) = Limit [(e^x){(1 + h) - 1}]/h h->0 = Limit [e^x(h/h)] h->0 But, h/h = 1, so: 8) f'(x) = Limit [e^x(1)] h->0 = e^x With Best Regards, Doctor Fwg, The Math Forum http://mathforum.org/dr.math/ |
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