The derivative of $${\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$$    with respect to $${\cos ^{ - 1}}\left[ {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right]$$    is equal to :

Solution :
$${\text{Let }}s = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,{\text{and }}t = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$$
we have to find out $$\frac{{ds}}{{dt}}$$  ; Putting $$x = \tan \,\theta ,$$   we get
$$\eqalign{ & s = {\sin ^{ - 1}}\left[ {\frac{{2\,\tan \,\theta }}{{1 + {{\tan }^2}\theta }}} \right] \cr & = {\sin ^{ - 1}}\left( {\sin \,2\theta } \right) \cr & = 2\theta \cr & = 2\,{\tan ^{ - 1}}x \cr & \therefore \,\frac{{ds}}{{dx}} = \frac{{2x}}{{1 + {x^2}}} \cr & {\text{and }}t = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) \cr & = {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) \cr & = {\cos ^{ - 1}}\left( {\cos \,2\theta } \right) \cr & = 2\,\theta \cr & = 2\,{\tan ^{ - 1}}x \cr & \therefore \,\frac{{dt}}{{dx}} = \frac{2}{{1 + {x^2}}} \cr & \therefore \,\frac{{ds}}{{dt}} = \frac{{\frac{{ds}}{{dx}}}}{{\frac{{dt}}{{dx}}}} = \frac{2}{{1 + {x^2}}} \times \frac{{1 + {x^2}}}{2} = 1 \cr} $$