Proposition 1

For every number $a > 0$ and $\neq 1$ the exponential function $a^\lambda$ admits at every point $x \in \mathbf{R}$ a derivative equal to $(\log_e a)a^\lambda$ where e is a number $> 1$ (independent of $a$).

Applying th. 1 to the case where E is the field $\mathbf{R}$ itself now shows that $a^\lambda$ has a derivative equal to $\varphi(a).a^\lambda$ at every point, where $\varphi(a)$ is a real number $\neq 0$ depending only on $a$. Let $b$ be a second number $> 0$ and $\neq 1$; the function $b^\lambda$ has a derivative equal to $\varphi(b).b^\lambda$ from the above; on the other hand, we have $b^\lambda = a^\lambda \log_a b$ so (I, p. 9, prop. 5) the derivative of $b^\lambda$ is equal to $\log_a b.\varphi(a)b^\lambda$; on comparing these two expressions we obtain

(2)

One deduces that there is one unique number $b$ such that $\varphi(b) = 1$; by (2) this relation is equivalent to $b = a^{1/\varphi(a)}$. It is conventional to denote the real number so obtained by e; from (2) one has $\varphi(a) = \log_e a$, which completes the proof of prop. 1.

One often writes $\exp x$ instead of $e^x$.

The definition of the number e shows that

(3)

which proves that $e^\lambda$ is strictly increasing, hence that $e > 1$.

In §2 (III, p. 105) we shall see how to calculate arbitrarily close approximations to e.