Prove that ln2 is the smallest value of the function f(x)=ln[1+square root(1+x^2)].

To determine the extreme value of a function we'll have
to determine the critical point of the function. The critical point of the function is
the root of the first derivative of the function.

We'll
have to calculate the first derivative of the given function. Since f(x) is a composed
function, we'll apply the chain rule.

f'(x) =
ln[1+sqrt(1+x^2)]'

f'(x) =
[2x/2sqrt(1+x^2)]/[1+sqrt(1+x^2)]

We'll simplify and we'll
get:

f'(x) =
x/sqrt(1+x^2)*[1+sqrt(1+x^2)]

We'll put f'(x) =
0

Since sqrt(1+x^2)>0, only the numerator could be
zero.

x = 0

The critical point
of the function is x = 0.

The minimumvalue of
the function is:

f(0)
= ln[1+sqrt(1+0^2)]

f(0)
= ln(1+1)

f(0) = ln2 
q.e.d.