Prove that the exponential function is differentiable

Imagine that you are writing a book on the foundations of analysis.

You have already proved that for each $a > 1$ there is a unique function $f_a(x) = a^x$ satisfying the following:

  1. $f_a$ is an isomorphism of ordered groups between $(\mathbb{R},+)$ and $(\mathbb{R}_{+},\cdot)$;
  2. $f_a(1) = a$.

It follows from the monotonicity and bijectivity of $f_a$ that it is continuous.

Now you would like to prove that $f_a$ is differentiable. At this point, you don’t know anything about integration, differential equations or power series.

What is the simplest or most elegant way of doing this?

Solutions Collecting From Web of "Prove that the exponential function is differentiable"

Since $f_a$ is a homomorphism, you only need to show differentiability at $0$, for

$$\frac{f_a(x+h) – f_a(x)}{h} = f_a(x)\frac{f_a(h)-1}{h}.$$

Since $f_a$ is convex [you need to show that, of course], you know that

$$\frac{f_a(h) – 1}{h}$$

is monotonically increasing in $h\in \mathbb{R}\setminus \{0\}$, hence the one-sided derivatives

$$D^+f_a(0) = \lim_{h \searrow 0}\frac{f(h)-1}{h},\quad D^-f_a(0) = \lim_{h\nearrow 0} \frac{f_a(h)-1}{h}$$

exist both. So it remains to see that they are equal. But since $f_a$ is a homomorphism we have

D^-f_a(0) &= \lim_{h\searrow 0}\frac{f_a(-h)-1}{-h} = \lim_{h\searrow 0} \frac{\frac{1}{f_a(h)}-1}{-h}\\
&= \lim_{h\searrow 0}\frac{1}{f_a(h)}\cdot\frac{1-f_a(h)}{-h}\\
&= \lim_{h\searrow 0}\underbrace{\frac{1}{f_a(h)}}_{\to 1} \cdot \underbrace{\frac{f_a(h)-1}{h}}_{\to D^{+}f_a(0)}\\
&= D^+f_a(0).