Mathematics · 2020
JEE Main · 5 September 2020, Shift 2 · Q59
The derivative of tan^-1((√(1+x^2)-1)/x) with respect to tan^-1((2 x √(1-x^2))/(1-2 x^2)) at x=1/2 is:
The derivative of $\displaystyle \tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\displaystyle \tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)$ at $\displaystyle x=\frac{1}{2}$ is :
Official answer
(3)
$\displaystyle \frac{\sqrt{3}}{10}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.