How does $f(x)= x \sin(\frac{\pi}{x})$ behave? I think this function is increasing for $x>1$ but wanted to find the reason. So I thought about taking the derivative:
$f(x)= x \sin(\frac{\pi}{x})$
Aplying the chain an the product rule, we get:
$f'(x)= \sin(\frac{\pi}{x})-\frac{\pi}{x} \cos (\frac{\pi}{x})$
The function is increasing if the derative is more than or equal to $0$, so:
$\sin(\frac{\pi}{x})-\frac{\pi}{x} \cos (\frac{\pi}{x}) \ge 0$
$\sin(\frac{\pi}{x}) \ge \frac{\pi}{x} \cos (\frac{\pi}{x}) $
Since $ \cos ( x) > 0$, if $ 0< x < \pi$,  $ \cos (\frac{\pi}{x}) > 0 $, because $ 0<\frac { \pi}{x}< \pi$.
$ \tan (\frac{\pi}{x}) \ge \frac{\pi}{x}$
I get to this point and don't know how to continue. I'd like you to help me or give me a hint, or maybe see a different way of showing it. Anyway, thanks.

Try the Maclaurin series for tangent:
$$\tan x = x + \frac{x^3}{3} + \frac{2 x^5}{15} + \ldots$$
All terms are positive when $x > 0$, so $\tan x > x$. This proves that $\tan \frac{\pi}{x} > \frac{\pi}{x}$ for $0 < x < \frac{\pi}2$.
Alternatively, define $f(x) = \tan x - x$. Then $f(0) = 0$ and $f'(x) = \sec^2 x - 1 > 0$ for $0 < x < \frac{\pi}2$, so $f(x)$ is increasing, and $f(x) > 0$ on $0 < x < \frac{\pi}2$.
Now, here's the catch: remember that tangent is not defined at $x = \frac{\pi}2$. However, the argument of your tangent function is $\frac{\pi}{x}$. So, for $x > 2$, this fraction decreases from $x = 2$ to $x = \infty$, and in fact, approaches zero. Therefore, you are safe with your argument showing that the original function is increasing for $x > 2$. For $1 < x \leq 2$, you need to be a bit more careful.
