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MHT CET · Maths · Inverse Trigonometric Functions

The derivative of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) w.r.t. \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)\) at \(x=0\) is

  1. A \(\frac{1}{8}\)
  2. B \(\frac{1}{4}\)
  3. C \(\frac{1}{2}\)
  4. D \(1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{4}\)

Step-by-step Solution

Detailed explanation

Let \(u = \tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\). Substitute \(x = \tan\theta\), then \(u = \frac{1}{2}\tan^{-1}x\). \(\frac{du}{dx} = \frac{1}{2(1+x^2)}\). At \(x=0\), \(\frac{du}{dx} = \frac{1}{2}\).
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