jee-main 2020 Q64

jee-main · India · session2_05sep_shift2 Differentiating Transcendental Functions Evaluate derivative at a point or find tangent slope

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) $\frac{\sqrt{3}}{10}$

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) $\frac{\sqrt{3}}{10}$

Paper Questions

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