The derivative of \(\sin^{-1}\left(\displaystyle\frac{2x}{1+x^2}\right)\) w. r. t. \(\tan^{-1}\left(\displaystyle\frac{2x}{1-x^2}\right)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1
Let
\( \displaystyle u = \sin^{-1}\left(\frac{2x}{1+x^2}\right), \quad v = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \)
The substitution x = tan θ simplifies both. For −1 < x < 1, take \(-\frac{\pi}{4} \lt \theta \lt \frac{\pi}{4}\), so that 2θ lies between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\):
\( \displaystyle \frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta \)
\( \displaystyle \frac{2x}{1-x^2} = \frac{2\tan\theta}{1-\tan^2\theta} = \tan 2\theta \)
Hence
\(u = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}x\)
\(v = \tan^{-1}(\tan 2\theta) = 2\theta = 2\tan^{-1}x\)
So u = v, and the derivative of u with respect to v is
\( \displaystyle \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2/(1+x^2)}{2/(1+x^2)} = 1 \)