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AP EAMCET · Maths · Differentiation

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

  1. A \(\frac{2}{\left(1+x^2\right)}\)
  2. B \(\frac{1}{2\left(1+x^2\right)}\)
  3. C \(\left(1+x^2\right)\)
  4. D \(2\left(1+x^2\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{2\left(1+x^2\right)}\)

Step-by-step Solution

Detailed explanation

Given, \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) Let, \(x=\tan \theta\) So, \(y=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)\)…