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

Calculation

Lets \(s = \sin^{-1}(\frac{2x}{1+x^2})\) and \(t = \cos^{-1}(\frac{1-x^2}{1+x^2})\) 

We have to find out

\(\frac{ds}{dt}\)

Putting x = tan θ, we get

\(s = \sin^{-1}[\frac{2 \tan θ}{1+\tan^2 θ}] = \sin^{-1}(\sin 2θ) = 2θ = 2 \tan^{-1} x\)  

∴ \(\frac{ds}{dx} = \frac{2}{1+x^2}\)  and \(t = \cos^{-1}(\frac{1-x^2}{1+x^2}) = \cos^{-1}(\frac{1-\tan^2 θ}{1+\tan^2 θ})\)

= cos^-1(cos 2θ) = 2θ = 2 tan^-1x

∴ \(\frac{dt}{dx} = \frac{2}{1+x^2}\)

∴ \(\frac{ds}{dt} = \frac{ds/dx}{dt/dx} = \frac{2}{1+x^2} \times \frac{1+x^2}{2} = 1\)

Hence option 1 is correct