Derivative of e^x

Usually, the first thing about e that one learns in school is the definition e=lim n→∞ (1+1/n)^n. Later on one is taught if f(x)=e^x then f'(x)=e^x. Most school textbooks on mathematics however lack proof for this, even though it is quite simple.

From the definintion of the derivative: f'(x)=(f(x+Δx)-f(x))/Δx when Δx→0
f'(x)=(e^(x+Δx)-e^x)/Δx
f'(x)=(e^x*e^Δx-e^x)/Δx
f'(x)=e^x*(e^Δx-1)/Δx

Now, all that remain is to show that (e^Δx-1)/Δx→1 as Δx→0
To do this one inserts e=(1+1/n)^n when n→∞.
f'(x)=e^x*((1+1/n)^n*Δx-1)/dx
As n→∞ and Δx→0, we can get rid of one of the limits by writing n=1/Δx. This makes the above equation simplify to
f'(x)=e^x*((1+Δx)^1 - 1)/Δx
f'(x)=e^x*Δx/Δx=e^x

There is a simple reason why one rarely sees a proof of the value of the derivative of e^x in textbooks. One never starts from e=2.718... or as the value of some limit and prove things from there on. It doesn't make sense, it would be like defining π from one of the many series that converge to it (or to some number related to it). Defining e like that doesn't explain why we have given a short and convenient name to this number and not to the value of some other limit.

Another flaw is that one has to give a meaning to things like e^π. The very notion of powers is tied to the exponential function, and this is where one should start, with a function. One usually starts by defining exp(x) to be the function satisfying the differential equation y'=y and such that exp(0)=1.

Calculating exp(x)

If you know about Taylor series, or more generally about expanding a function as a series, you will know that we can write:

exp(x)=1+exp'(0)x + o(x)=1+x + o(x)

Details on what o(x) means are here, but basically it means "something small compared to x".

We can integrate this series to get:

exp(x)=x+x²/2+o(x^2)+c

where c is a constant of integration. One of the properties of such series is that when you add on terms, the first ones stay the same, so c=1

We can continue this, and so we have

exp(x)=1+x+x²/2+x³/6+...+xⁿ/n!

which we can write as exp(x)=Σ^∞_i=0xⁿ/n!

A bit of analysis shows that this series converges everywhere, and thus we can calculate exp(x) everywhere in this manner.

Fundamental properties of the exponential

Here we will prove the 2 basic properties:

• exp(x+y)=exp(x)*exp(y)
• exp(xy)=(exp(x))^y

Consider the function f(x)=exp(x+y).
We have, f(0)=exp(y), and f'=f so f'(0)=exp(y). We use the same method for calculating f(x) as we did for calculating exp(x):

f(x)=exp(y)+exp(y)x +o(x)

We do our integration and obtain

f(x)=exp(y)+exp(y)x+exp(y)x²/2 +o(x²)

Now this is just exp(y) times the beginning of the series for exp(x). Since integration is linear, we can in fact pull out the exp(y) factor, and have exp(x+y)=exp(y)*exp(x)

Before you know about exponentials, it only makes sense to define integer powers. We note that

exp(1*x)=exp(x)
exp(2*x)=exp(x+x)=(exp(x))2
exp(n*x)=exp(x+x+...+x)=(exp(x))n

So the 2 expressions coincide for integer values of y. Until now we haven't defined what it meant for a number to be raised to the √2 th power for example. This means we are able to define exp(yx)=(exp(x))^y for all values of y, we extend the meaning of "raise to a power"

You may notice that the exponential function is behaving in the same way as a power function, with the base exp(1). We then have exp(x)=e^x.

It is left as an exercise to the reader to deduce from this other properties of the exponential, such as its inverse function, or that exp(1)= lim_{n→ ∞}(1+1/n)ⁿ. The derivative of the exponential function being equal to the function itself is really much more of a defining feature than a property.