Let #f(x)=cos^(-1)(2x^2-1)# and #g(x)=sin^(-1)(1-x^2)#
what we are seeking is #(df)/(dg)#, which is equal to #((df)/(dx))/((dg)/(dx))#
As #f=cos^(-1)(2x^2-1)#, we have #cosf=2x^2-1#
and differentiating #-sinf*(df)/(dx)=4x#
or #(df)/(dx)=-(4x)/sinf=-(4x)/sqrt(1-cos^2f)=-(4x)/sqrt(1-(2x^2-1))^2#
= #-(4x)/sqrt(1-4x^4+4x^2-1)=-(4x)/(2xsqrt(1-x^2))=-2/sqrt(1-x^2)#
Similarly we have #sing=1-x^2# and #cosg*(dg)/(dx)=-2x#
or #(dg)/(dx)=-(2x)/cosg=-(2x)/sqrt(1-sin^2g)=-(2x)/sqrt(1-(1-x^2)^2)#
= #-(2x)/sqrt(1-1+2x^2-x^4)=-2/sqrt(2-x^2)#
Hence #(df)/(dg)=(-2/sqrt(1-x^2))/(-2/sqrt(2-x^2))#
= #sqrt((2-x^2)/(1-x^2))#
