Q103 / 160 · TS EAMCET 2015 · Mathematics

If $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$ then $\frac{d y}{d x}$ is equal to

  • A
    $\frac{2}{1+x^2}$
  • B
    $\frac{4}{1+x^2}$
  • C
    $\frac{6}{1+x^2}$
  • Correct answer
    D
    $\frac{7}{1+x^2}$

Solution

We have, $$ y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right) $$ Put $x=\tan \theta$, we get $$ \begin{aligned} & y=\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right) \\ & +\tan ^{-1}\left(\frac{4 \tan \theta-4 \tan ^3 \theta}{1-6 \tan ^2 \theta+\tan ^4 \theta}\right) \\ & =\tan ^{-1} \tan 3 \theta+\tan ^{-1} \tan 4 \theta \\ & =3 \theta+4 \theta \\ & \Rightarrow \quad y=7 \theta \\ & \therefore \quad y=7 \tan ^{-1} x \\ & \end{aligned} $$ On differentiating both sides w.r.t. $x$, we get $$ \frac{d y}{d x}=\frac{7}{1+x^2} $$

Hence, the correct answer is option: 'D'.

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