To solve this problem, we must differentiate each of the functions in List I and match them with the corresponding derivatives provided in List II.
- Derivative \(A\): \( \frac{d}{dx}[tan^{-1}(\frac{3x-x^3}{1-3x^2})] \)
- The expression inside \(tan^{-1}\) is of the form \( \frac{u}{v} \) where \( u = 3x-x^3 \) and \( v = 1-3x^2 \).
- Using the derivative rule for \(tan^{-1}(x)\), which is \(\frac{1}{1+x^2}\), and chain rule, the derivative is \(\frac{1}{1+\left(\frac{3x-x^3}{1-3x^2}\right)^2} \times \frac{d}{dx}(\frac{3x-x^3}{1-3x^2})\).
- Upon differentiating and simplifying, we find it matches option \(I\): \(\frac{3}{1+x^2}\).
- Derivative \(B\): \( \frac{d}{dx}[cos^{-1}(\frac{1-x^2}{1+x^2})] \)
- The given function resembles a standard transformation for trigonometric functions.
- The derivative of \(cos^{-1}(x)\) is \(-\frac{1}{\sqrt{1-x^2}}\), but here simplify the expression using trigonometric identities, leading us to \(-\frac{2}{1+x^2}\).
- This matches with option \(IV\).
- Derivative \(C\): \( \frac{d}{dx}[cos^{-1}(\frac{2x}{1+x^2})] \)
- The derivative follows a similar process as \(B\), when considering inverse trigonometric transformations.
- After applying the chain rule, we find \(-\frac{2}{1+x^2}\).
- Thus it matches option \(III\).
- Derivative \(D\): \( \frac{d}{dx}[cot^{-1}(\frac{3x-x^3}{1-3x^2})] \)
- The expression is similar to \(A\), but differentiated with respect to \(cot^{-1}(x)\), which gives us \(-\frac{1}{1+x^2}.\)
- Simplifying confirms that the derivative is \(-\frac{3}{1+x^2}\).
- Thus it corresponds to option \(II\).
The correct matching is therefore: A-I, B-IV, C-III, D-II.