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?

Solution 1:

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

\begin{align} 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). \end{align}