JEE Main 2020 (Online) 5th September Evening ShiftQuestion 1
The derivative of $\tan^{-1}\left(\dfrac{\sqrt{1+x^{2}}-1}{x}\right)$ with respect to $\tan^{-1}\left(\dfrac{2x\sqrt{1-x^{2}}}{1-2x^{2}}\right)$ at $x=\dfrac{1}{2}$ is :
Differentiation and its Application
Solution
Concept. Simplify each function by a substitution, then divide their derivatives. Formulas used. (1) $x=\tan\theta$: $\dfrac{\sec\theta-1}{\tan\theta}=\tan\dfrac\theta2$, so $u=\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)=\dfrac12\tan^{-1}x$(2) $x=\sin\phi$: $\dfrac{2x\sqrt{1-x^2}}{1-2x^2}=\tan2\phi$, so near $x=\frac12$, $v=2\sin^{-1}x$Step 1 - Derivatives. By (1), (2): $$\frac{du}{dx}=\frac{1}{2(1+x^2)},\qquad\frac{dv}{dx}=\frac{2}{\sqrt{1-x^2}}.$$ (At $x=\frac12$, $2\phi=\frac\pi3$ lies in $\left(-\frac\pi2,\frac\pi2\right)$, so (2) holds near this point.) Step 2 - At $x=\frac12$. $\frac{du}{dx}=\frac25$ and $\frac{dv}{dx}=\frac{4}{\sqrt3}$. Step 3 - Ratio. $\dfrac{du}{dv}=\dfrac{2/5}{4/\sqrt3}=\dfrac{\sqrt3}{10}$. Answer: (C)