If \(f(x) = \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)\), then \(f'\left(\dfrac{1}{2}\right) = \)

Given \(f(x) = \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)\)

Let \(x = \tan \theta\), then \(\theta = \tan^{-1}x\).

Substituting \(x = \tan \theta\) in the given function:

\(f(x) = \sin^{-1}\left(\dfrac{2\tan\theta}{1 + \tan^2\theta}\right)\)

\(f(x) = \sin^{-1}(\sin 2\theta)\)

For \(x = \dfrac{1}{2}\), \(\theta = \tan^{-1}\left(\dfrac{1}{2}\right)\), which means \(2\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\).

Therefore, \(\sin^{-1}(\sin 2\theta) = 2\theta\).

\(f(x) = 2\tan^{-1}x\)

Differentiating with respect to \(x\):

\(f'(x) = \dfrac{2}{1 + x^2}\)

Substituting \(x = \dfrac{1}{2}\):

\(f'\left(\dfrac{1}{2}\right) = \dfrac{2}{1 + \left(\dfrac{1}{2}\right)^2} = \dfrac{2}{1 + \dfrac{1}{4}} = \dfrac{2}{\dfrac{5}{4}} = \dfrac{8}{5}\)

Answer: \(\dfrac{8}{5}\)