JEE PYQ: Differentiation Question 7
Question 7 - 2020 (05 Sep Shift 2)
The derivative of $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ at $x = \frac{1}{2}$ is:
(1) $\frac{2\sqrt{3}}{5}$ (2) $\frac{\sqrt{3}}{12}$ (3) $\frac{2\sqrt{3}}{3}$ (4) $\frac{\sqrt{3}}{10}$
Show Answer
Answer: (4) $\frac{\sqrt{3}}{10}$
Solution
$u = \frac{1}{2}\tan^{-1}x$, $v = 2\sin^{-1}x$. $\frac{du}{dv} = \frac{\sqrt{1-x^2}}{4(1+x^2)}$. At $x = \frac{1}{2}$: $\frac{\sqrt{3}}{10}$.
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