If $f(x)=2 \tan ^{-1} x+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then

$f^{\prime}(x)=0$ for all $x<0$

We know that

$\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\left\{\begin{array}{cl} 2 \tan ^{-1} x, & x \geq 0 \\ -2 \tan ^{-1} x, & x \leq 0 \end{array}\right.$

∴  $f(x)=\left\{\begin{array}{cl} 4 \tan ^{-1} x, & x \geq 0 \\ 0, & x \leq 0 \end{array}\right.$

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