Using economics to understand informally why exp'(x) = exp(x).
Notes by Martin Flashman

Recall that there is a relation between the exponential function with the base e and continuously compounded interest.

This is based on the characterization  of e by the estimation with [1+1/u]u when u is a large number.

The economic context: I invest an amount (the principle, P) to be paid interest at an annual rate of r = 100r % . The interest could be
  • paid only at the date of withdrawal (called simple interest), or 
  • compounded n times a year, or 
  • compounded continuously.
I would like to know in each case:
How much will
I have after t years where t can be any reasonable part of a year?
Analysis 1: For interest paid only at the date of withdrawal. The value S of a principle P with interest paid only at the date of withdrawal t part of a year is given by
S(t)= P + Prt = P(1+rt). (1)
Notice: This is a linear function of t,  with slope or rate Pr.
Analysis 2: For interest compounded n times a year.
The value A of a principle P compounded n times annually at an annual rate of r for t = k /n parts of a year is given by
A(t)= P*(1+r/n)k = P*(1+r/n)nt. (2)
Notice: This is a discrete exponential function of k or t , with base b=(1+r/n).
The increment in value for the first period is precisely Pr/n.
So the rate for the first period of compounding is (Pr/n) / (1/n) =Pr.

Analysis 3: For interest compounded continuously.

We use the previous analysis to estimate. The value A of a principle P compounded n times annually at an annual rate of r for t = k /n parts of a year was given by
A(t)= P*(1+r/n)k. (2)

Letting u = n/r , we have 1/u = r/n.

Note that k = t n
  and we have
A(t)= P (1+ 1/u)urt = P*((1+ 1/u)u)rt .     (2*)

By considering compounding a large number of times per year, we estimate continuously compounding interest.

When n is large, so is u,
and we have A(t) is approximately P*
ert.  This gives the value for the limiting continously compounded value:
A(t) = P* ert .   (3)

This is a continuous exponential function of t, with base e.
The increment in value for the first very short period of t = 1/n  with n large is approximately Pr/n = Prt.
So the rate for the first period of compounding is approximately
(Pr/n) / (1/n) = Pr.  (4)
This last statement says the rate for any initial short period is approximately the same and equal to the simple (linear) interest rate of case 1!

We can now simplify this by assuming that I invested the value of 1 unit.
So in the last case at a rate of r per year compounded continously, the value at time t years is given by
A(t) = ert .

Note: When r = 1, the formula tells us that for continously compounding A is the natural exponential function, A(t) = exp(t) =
et .
Now for the calculus:

First, we notice that for any exponential function f (x) = b
t ,
f ' (x) = f (x)* f '(0).
This is a direct result of the fact about exponents:
[bx+h - bx ]/h = bx * [bh - 1]/h             [f (0) = 1].

Notice also that we can intepret f ' (0) as the slope of the line tangent to the graph of f  at x= 0.
So, to show that exp'(x)=exp(x), we need only show that exp'(0)=1.

Thus, considering the definition of the derivative,
we need to find the limit as h approaches 0 for
[eh - 1]/h.

Now we turn to our economic analysis for help.
We think of the context of continuously compounding with a principal of P = 1 and rate r = 1.
Thus A(t) =
et .
For a small
value of h, use h = 1/n where n is a large natural number.
The economic estimates
for the rate of change for the first period of compounding correspond precisely to the estimation of [eh - 1]/h.
But the economic interpretation said that for any estimation of h this rate would be the same, that is, the rate is Pr = 1.
So when we consider these estimates we conclude that exp'(0) = 1 and exp'(x) = exp(x).

An alternative approach to showing exp'(0)= 1:
 Since  e is estimated by 
(1+ 1/n)n for large values of n, choose h = 1/n for a large value of n.
[eh - 1]/h is approximately
[((1+ 1/n)n)h - 1]/h = [((1+ 1/n)nh - 1]/h = [((1+ 1/n)n)h - 1]/h = [(1+h) - 1]/h = h/h =1!
Thus with n  getting larger, h gets closer to 0 and the estimate is closer to the derivative exp'(0).
So, exp'(0) = 1.

Notice that the same reasoning applied to the estimated rate in (4) can be used to show that
for f (x) = exp(rt),  f  '(x) = r
exp(rt).[ Or this result follows from the chain rule!]