Mean value theorem with trigonometric functions Let $f(x) = 2\arctan(x) + \arcsin\left(\frac{2x}{1+x^2}\right)$


*

*Show that $f(x)$ is defined for every $ x\ge 1$

*Calculate $f'(x)$ within this range

*Conclude that $f(x) = \pi$ for every $ x\ge 1$
Can I get some hints how to start? I don't know how to start proving that  $f(x)$ is defined for every $ x\ge 1$ and I even had problems calculating $f'(x)$
Thanks everyone!
 A: The expression for $f(x)$ is defined whenever
$$
-1\le\frac{2x}{1+x^2}\le1
$$
that is
\begin{cases}
2x\le 1+x^2\\
2x\ge -(1+x^2)
\end{cases}
or
\begin{cases}
x^2-2x+1\ge0\\
x^2+2x+1\ge0
\end{cases}
which is satisfied for every $x$.
Computing the derivative of
$$
g(x)=\arcsin\frac{2x}{1+x^2}
$$
just requires some patience; consider $h(x)=2x/(1+x^2)$; then
$$
g'(x)=\frac{1}{\sqrt{1-h(x)^2}}h'(x)
$$
by the chain rule. Now
$$
1-h(x)^2=1-\frac{4x^2}{(1+x^2)^2}=\frac{1+2x^2+x^4-4x^2}{(1+x^2)^2}=
\frac{(1-x^2)^2}{(1+x^2)^2}
$$
so
$$
\frac{1}{\sqrt{1-h(x)^2}}=\frac{1+x^2}{|1-x^2|}
$$
Note the absolute value! Always remember that $\sqrt{a^2}=|a|$. On the other hand, we can write $\sqrt{(1+x^2)^2}=1+x^2$, because this expression is always positive.
Now we tackle $h'(x)$:
$$
h'(x)=2\frac{1\cdot(1+x^2)-x\cdot 2x}{(1+x^2)^2}=\frac{2(1-x^2)}{(1+x^2)^2}
$$
and so, putting all together, we have
$$
g'(x)=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}=
\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}
$$
Note that this is not defined for $x=1$ or $x=-1$.
Thus you have
$$
f'(x)=\frac{2}{1+x^2}+\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}
$$
For $x>1$ we have $|1-x^2|=x^2-1$, so $\frac{1-x^2}{|1-x^2|}=-1$
and therefore
$$
f'(x)=0
$$
for $x>1$ and the function is constant in the interval $(1,\infty)$. Since it is continuous at $1$ it is constant also on $[1,\infty)$. The constant value is
$$
f(1)=2\arctan1+\arcsin1=2\frac{\pi}{4}+\frac{\pi}{2}=\pi
$$
Note also that, for $x<-1$, we can draw the same conclusion and
$$
f(x)=-\pi \qquad(x\le -1)
$$
On the other hand, for $-1<x<1$ we have
$$
f'(x)=\frac{4}{1+x^2}
$$
so
$$
f(x)=k+4\arctan x
$$
(because in $(-1,1)$ the two functions have the same derivative), where $k$ is constant; this can be evaluated by seeing that $f(0)=0$, so $k=0$.
Thus your function $f$ can also be represented by
$$
f(x)=\begin{cases}
-\pi & \text{for $x\le-1$}\\
4\arctan x & \text{for $-1<x<1$}\\
\pi & \text{for $x\ge 1$}
\end{cases}
$$
Note. I showed the full exercise because it uses just basic facts. I wanted to underline how splitting the computation into pieces can lead to the result in an easier way than by having gigantic expressions to deal with.
Here's a picture of the graph.

A: Hint
Recall that the $\arctan$ function is defined on $\Bbb R$ while the $\arcsin$ function is defined on $[-1,1]$. Compute the derivative $f'(x)$ and prove that it's equal to $0$. Conclude that $f$ is a constant which we can determinate by taking the limit of $f$ at $+\infty$ 
