#2015_13 I just started learning calculus

First note that it is valid to consider this infinite tetration in the neighborhood of $x = \frac{\pi}{4},$ and in this neighborhood $y = f(x)$ is both continuous and differentiable.

As such, we then observe that, (since $y \gt 0$ in this neighborhood),

$\ln(y) = y*\ln(\tan(x)) \Longrightarrow \dfrac{\ln(y)}{y} = \ln(\tan(x)).$

Differentiating both sides (using the quotient and chain rules) we have that

$\dfrac{1 - \ln(y)}{y^{2}} * \dfrac{dy}{dx} = \dfrac{\sec^{2}(x)}{\tan(x)} \Longrightarrow \dfrac{dy}{dx} = \dfrac{y^{2}\sec(x)\csc(x)}{1 - \ln(y)}.$

Now at $x = \frac{\pi}{4} \Longrightarrow \tan(x) = 1$ we have that

$y(1) = 1$ and $\sec(x) = \csc(x) =\sqrt{2},$ thus

$f'(\frac{\pi}{4}) = \dfrac{1*2}{1 - 0} = \boxed{2}.$