I won't swear to this but here goes

$v(x)=\tan^{-1}\left(\dfrac {2x}{1-x^2}\right)$

$u(x)=\tan^{-1}(x)$

$x(u)=\tan(u)$

$\dfrac {dv}{du} = \dfrac {dv}{dx} \dfrac{dx}{du}$

$\dfrac{dv}{dx}=\dfrac{2}{1+x^2}$

$\dfrac{dx}{du}=\sec^2(u)=\sec^2(\tan^{-1}(x))=1+x^2$

$\dfrac {dv}{du} = \dfrac {dv}{dx} \dfrac{dx}{du}=\dfrac{2}{1+x^2}\cdot(1+x^2)=2$

so

$\dfrac{d\left(\tan^{-1}\left(\dfrac {2x}{1-x^2}\right)\right)}{d\left(\tan^{-1}(x)\right)}=2$