Derivative of an Exponential

[weberju]

I was wondering how one learns how to do the derivative of an exponential. Before we learn all the rules for derivatives, we use the definition of the derivative via the limit to find the rules that everyone uses (i.e derivative of x^n = n*x^(n-1)). But I can't figure out how to apply the definition of the derivative to the exponential without using the Taylor series expansion (which uses derivatives in it's definition). This therefore lock's me into a paradixical loop. So basically I have two questions.
1) Is there a way to expand functions without using derivatives?
2) Is the exponential just simply defined as: exp(x)=1+x+(1/2)x^2+..., which would make my question a silly one.

[OfficeShredder]

There are a number of ways to define e^x, the taylor series being one of them.

If you're defining it as the inverse of ln(x), try this:

lim h->0 (e^(x+h) - e^x)/h = lim e^x(e^h-1)/h = e^x lim(e^h-1)/h

Let h=lnt, so e^x lim t->1 (t-1)/lnt

ignore the e^x for now, let's show that limit is equal to 1. I'm assuming you've done ln(x) is the integral of 1/x for this point.

(t-1)/ln(t) = 1/[ln(t)/(t-1)] = 1/[ ln(t) - ln(1)/(t-1) ].

If you notice, once you take the limit of that as t approaches 1, you get 1/ln'(1) which is 1/(1/1) = 1. So the derivative turns out to be e^x * 1 = e^x

[HallsofIvy]

If you are defining e^x as the inverse function to y= ln x, then you don't need to go back to the original definition.

If y= e^x then x= ln y and dx/dy= 1/y. by the chain rule, then dy/dx= y= e^x.

If, on the other hand, you define a^x, a any positive real number, by:
1) a^n, for n a positive integer, is a multiplied by itself n times.

2) a^0= 1

3) If n is a negative integer then a^n= 1/a^(-n). Since -n is a positive integer, the denominator was defined in 1) above.

4) for n a positive integer, a^(1/n) is the positive th root of a.

5) a^(m/n)= (a^(1/n))^m where a^(1/n) is defined in 4)

That defines a^x for x any rational number. If x is an irrational number, then there exist a sequence {r_i} of rational numbers converging to x.
6) a^x= lim a^(r_i) (so that a^x is a continuous function.

It's easy to show that a^(x+h)- a^x= a^x(a^h- 1) so that the "difference quotient" (a^(x+h)- a^x)/h= a^x (a^h-1)/h, a^x itself times something that depends only on h. IF the limit (a^h-1)/h exists as h goes to 0, then the derivative of a^x exists and is a^x times a constant. We then DEFINE "e" as the value of a such that that constant is 1. The hard part is proving that the limit exists. That's why I prefer first defining ln x as "the integral of (1/t)dt from 1 to x" and then defining e^x as the inverse function to ln x.