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
- A \(\frac{1}{8}\)
- B \(\frac{1}{4}\)
- C \(\frac{1}{2}\)
- D \(1\)
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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