Let for all . By the chain and product rules,

on , so is constant. Since , we conclude

In particular is never , and

Now let
and define a function
by

We have

for all , so is constant, and . Thus

and by (12.24),

Next suppose you know some function such that for all and . (You do know such a function from your previous calculus course.) Let

Then by the product and chain rules,

so is constant on , and since , we have . By (12.24),

Now I will try to construct a function satisfying the differential equation (12.23) by hoping that is given by a power series. Suppose

Since , we must have , and

By the differentiation theorem,

and

By the differentiation theorem again,

so

Hence

Repeating the process, we get

so

I see a pattern here: .

Proof: It is clear that .
The formal derivative of is

so the [still unproved] differentiation theorem says that . It follows from our discussion above that is never ,

and

It is clear that is real for all . In fact, we must have for all , since is continuous (differentiable functions are continuous) and if for some , the intermediate value theorem would say for some between and . Since on , is strictly increasing on .

Use this to prove that

i.e.,

(Note that for , this says

Proof: The uniqueness of follows from the fact that is strictly increasing on . Let . From the expansion , we see that . Since is continuous, we can apply the intermediate value theorem to on to conclude for some . If , then , so for some , and where . Since , the theorem has been proved in all cases.