Given a curve $y$ = $\frac{1}{1+x^2}$ determine point $A$ on that curve such that tangent on curve $y$ through point $A$ makes largest angle with positive part of $x$-axis (in terms of absolute value)
Let $y = f(x)$, let coordinates of $A$ be $A(x_0, f(x_0))$ and let $\theta$ be angle we want to maximize.
Slope of a tangent line through point $A$ is $f'(x_0) = \frac{-2x_0}{(1+x_0^2)^2}$, so it means that $\tan \theta = \frac{-2x_0}{(1+x_0^2)^2}$, $\theta = arctan(\frac{-2x_0}{(1+x_0^2)^2})$. Now I need to find derivative of $arctan(\frac{-2x_0}{(1+x_0^2)^2})$, equalize that with $0$ to find extreme point, and prove that it's a maximum point. We have $(arctan(\frac{-2x_0}{(1+x_0^2)^2}))'$ = $\frac{4x_0^5-4x_0}{(1+x_0^2)^4 + 4x_0^4}$, from where I get, when I equalize with $0$, $x_0 = \pm1$. Point $x_0 = -1$ is a maximum point, but that isn't on postive part of $x$-axis. Where am I making the mistake?