Q60 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Mathematics

The derivate of \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) is equal to

  • A
    \(\frac{2}{\left(1+x^2\right)}\)
  • Correct answer
    B
    \(\frac{1}{2\left(1+x^2\right)}\)
  • C
    \(\left(1+x^2\right)\)
  • D
    \(2\left(1+x^2\right)\)

Solution

Given, \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) Let, \(x=\tan \theta\) So, \(y=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)\) \(\begin{aligned} & =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\ & =\tan ^{-1}(\tan \theta / 2)=\theta / 2 \\ \Rightarrow y & =\frac{1}{2} \tan ^{-1} x \Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1}{1+x^2}\right) \end{aligned}\) Hence, option (b) is correct.

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

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