Question

The derivative of $\tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right)$ w.r.t. $\tan^{-1} \left( \frac{2x\sqrt{1-x^2}}{1-2x^2} \right)$ at $x = 0$ is

• A. 1/8
• B. 1/4
• C. 1/2
• D. 1

Hint:

$\tan^{-1} \frac{\sqrt{1+x^2}-1}{x} = \frac{1}{2} \tan^{-1} x$.

Answer 1

The Correct Option is B

Step 1: Simplify first function
Let $u = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right)$. Put $x = \tan \theta$.
$u = \tan^{-1} (\frac{\sec \theta - 1}{\tan \theta}) = \tan^{-1} (\frac{1-\cos \theta}{\sin \theta}) = \tan^{-1} (\tan \frac{\theta}{2}) = \frac{1}{2} \tan^{-1} x$.
Step 2: Simplify second function
Let $v = \tan^{-1} \left( \frac{2x\sqrt{1-x^2}}{1-2x^2} \right)$. Put $x = \sin \phi$.
$v = \tan^{-1} (\frac{2 \sin \phi \cos \phi}{\cos 2\phi}) = \tan^{-1} (\tan 2\phi) = 2 \sin^{-1} x$.
Step 3: Differentiation
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{2(1+x^2)}}{\frac{2}{\sqrt{1-x^2}}}$.
At $x = 0$, $\frac{du}{dv} = \frac{1/2}{2} = \frac{1}{4}$.
Final Answer: (B)