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\[f(x)=\sin ^{-1} ( (2x) / (1+x^2) ) \]

\[g(x)=\cos ^{-1} ( (1-x^2) / (1+x^2) ) \]

but seems to lengthy to do. :( ?

df(x)/dx=2/(1+x^2) and dg(x)/dx=2/(1+x^2)
dx/dg(x)=(1+x^2)/2
so
df(x)/dg(x)=1

akanksha: what you mentioned before is right, but
(df(x)/dx)/ (dg(x)/dx) would be a better choice!

Ok right, you did the same thing. There is no other way to do it! :)

Cool ! Thanks.