It's a bit unfortunate I can't get latex2html working on my set-up so the equations here will be a bit unattractive.

We know the identity ln (e) = 1. However, have you ever wondered how e is defined?

(1) as x approaches 0, e = lim (1 + x) ^ (1/x)

(2) as x approaches infinity, e = lim (1 + 1/x)^x

Using these 2 basic definitions, we should be able to prove that d/dx (ln x) = 1/x.

We proceed as follows:

d/dx (ln x) = lim [ ln (x+d) - ln (x) ] / d : as d approaches zero

= lim [ ln { (x+d) / x } / d ] : as d approaches zero

= lim [ 1/d * ln { 1 + (d/x) } ] : as d approaches zero

= lim ln {1 + (d/x) } ^ (1/d) : as d approaches zero

Setting u = d/x, we have:

lim ln {1 + (d/x) } ^ (1/d) : as d approaches zero = lim ln (1 + u) ^ (1/u * 1/x) : as u approaches zero

= 1/x ln (1 + u) ^ (1/u)

= 1/x ln (e)

= 1/x

QED

We can now prove further d/dx (e^x) = e^x. Let y = e^x. Taking ln of both sides of the equation, we get

ln (y) = ln (e^x)

ln (y) = x ln (e)

ln (y) = x

By implicit differentiation, we get

y'/y = 1

y' = y

Substituting y with its original value, we get

y' = e^x

QED

We can also prove d/dx (a^x) = (a^x) ln (a). Let y = a^x. Taking ln of both sides of the equation, we get

ln (y) = ln (a^x)

ln (y) = x ln (a)

By implicit differentiation, we get

y'/y = ln (a)

y' = y ln (a)

Substituting y with its original value, we get

y' = (a^x) ln (a)

QED