Is the function arctanx covex or concave in the positive real set.

In order to prove that f(x) is concave or convex, we'll
have to calculate the second derivative of the function.

If
the second derivative is negative, then the graph of f(x) is
concave.

If the second derivative is positive, then
the graph of f(x) is convex.

For the beginning, we'll
calculate the first derivative of
f(x):

f'(x)=1/(1+x^2)

Now,
we'll calculate f"(x) of the expression arctan x, or we'll calculate the first
derivative of f'(x).

f"(x) =
[f'(x)]'

Since f'(x) is a ratio, we'll apply the quotient
rule:

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

We'll put 1' = 0 and
we'll remove the brackets:

f"(x)=
-2x/(1+x^2)^2

Because of the fact that denominator is
always positive, then the numerator will influence the
ratio.

Since numerator is negative over the interval
[0,infinite), f"(x)<0.

So, the
function arctan x is concave over the positive real set of
numbers.