Q103 / 180 · KCET 2023 · Mathematics

If $u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ then $\frac{d u}{d v}$ is

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

Solution

Here, $u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ $\left.u=2 \tan ^{-1} x \quad \because \because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]$ $\begin{aligned} & \Rightarrow \frac{d u}{d x}=2 \times \frac{1}{1+x^2} \\ & \text { and } v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\end{aligned}$ $\begin{aligned} y & =2 \tan ^{-1} x \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right] \\ \Rightarrow \quad & \quad \frac{d v}{d x}=\frac{2}{1+x^2}\end{aligned}$ From Eqs. (i) and (ii), we get $\frac{d v}{d x}=1$

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

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