Let $f(x)=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$. Then,

all the above

We know that

$\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)= \begin{cases}2 \tan ^{-1} x, & \text { if }-1 \leq x \leq 1 \\ \pi-2 \tan ^{-1} x, & \text { if } x>1 \\ -\pi-2 \tan ^{-1} x, & \text { if } x<-1\end{cases}$

∴   $f(x)=\left\{\begin{array}{cl}4 \tan ^{-1} x, & \text { if }-1 \leq x \leq 1 \\ \pi & , \text { if } x>1 \\ -\pi & , \text { if } x<-1\end{array}\right.$

$\Rightarrow f^{\prime}(x)=\left\{\begin{array}{cl}\frac{4}{1+x^2}, & \text { if }-1<x<1 \\ 0, & \text { if }|x|>1\end{array}\right.$

$\Rightarrow f^{\prime}(2)=f^{\prime}(3)=0$ and $f^{\prime}\left(\frac{1}{2}\right)=\frac{16}{5}$