Match List I with List II
| LIST I | LIST II | ||
| A. | $\frac{d}{dx}\left[tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)\right]$ | I. | $\frac{3}{1+x^2}$ |
| B. | $\frac{d}{dx}\left[cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$ | II. | $\frac{-3}{1+x^2}$ |
| C. | $\frac{d}{dx}\left[cos^{-1}\left(\frac{2x}{1+x^2}\right)\right]$ | III. | $\frac{-2}{1+x^2}$ |
| D. | $\frac{d}{dx}\left[cot^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)\right]$ | IV. | $\frac{2}{1+x^2}$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-I, B-IV, C-III, D-II
(A) $\frac{d}{dx}\left[tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)\right]=\frac{d}{dx}(3\tan^{-1}x)=\frac{3}{1+x^2}$ (I)
(B) $\frac{d}{dx}\left[cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]=\frac{d}{dx}(2\tan^{-1}x)=\frac{2}{1+x^2}$ (IV)
(C) $\frac{d}{dx}\left[cos^{-1}\left(\frac{2x}{1+x^2}\right)\right]=\frac{d}{dx}\left(\frac{π}{2}-\sin^{-1}\frac{2x}{1+x^2}\right)$
$=\frac{d}{dx}\left(\frac{π}{2}-2\tan^{-1}x\right)=\frac{-2}{1+x^2}$ (III)
(D) $\frac{d}{dx}\left[cot^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)\right]$
$\frac{d}{dx}\left(\frac{π}{2}-\tan^{-1}\frac{3x-x^3}{1+3x^2}\right)=\frac{d}{dx}\left(\frac{π}{2}-3\tan^{-1}x\right)=\frac{-3}{1+x^2}$ (II)