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: Understanding the Concept:
Substitution makes inverse trigonometric derivatives significantly simpler.
For $\sqrt{1+x^2}$, use $x = \tan\theta$. For $\sqrt{1-x^2}$, use $x = \sin\phi$.
Step 2: Key Formula or Approach:
$u = \tan^{-1} (\frac{\sec\theta-1}{\tan\theta}) = \tan^{-1}(\tan \theta/2) = \frac{1}{2}\tan^{-1}x$.
$v = \tan^{-1} (\frac{2\sin\phi \cos\phi}{\cos 2\phi}) = \tan^{-1}(\tan 2\phi) = 2\sin^{-1}x$.
Step 3: Detailed Explanation:
$\frac{du}{dx} = \frac{1}{2(1+x^2)}$.
$\frac{dv}{dx} = \frac{2}{\sqrt{1-x^2}}$.
At $x=0$, $\frac{du}{dv} = \frac{1/2}{2} = 1/4$.
Step 4: Final Answer:
The value is 1/4.