AP EAMCET · Maths · Differentiation
The derivate of \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) is equal to
- A \(\frac{2}{\left(1+x^2\right)}\)
- B \(\frac{1}{2\left(1+x^2\right)}\)
- C \(\left(1+x^2\right)\)
- D \(2\left(1+x^2\right)\)
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, \(y=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)\)…
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