Question

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 \(x=0\), is

• A. \(\frac{1}{8}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{2}\)
• D. 1

Hint:

Use trigonometric substitutions to simplify inverse trigonometric expressions.

Answer 1

The Correct Option is B

To find 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 \(x=0\), we can use the concept of implicit differentiation. Let's go through the solution step-by-step.

1. Define \(y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) and \(z = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)\).
2. We need to calculate \(\frac{dy}{dz}\) at \(x=0\). By the chain rule, \(\frac{dy}{dz} = \frac{dy/dx}{dz/dx}\).
3. Start by finding \(\frac{dy}{dx}\): 
   \(y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\)

Using the chain rule and quotient rule:

1. \(\frac{dy}{dx} = \frac{1}{1 + \left(\frac{\sqrt{1+x^2}-1}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\)

Find the derivative of the quotient:

1. \(\frac{d}{dx}\left(\frac{\sqrt{1+x^2}-1}{x}\right) = \frac{x \cdot \frac{x}{\sqrt{1+x^2}} - (\sqrt{1+x^2}-1)}{x^2}\)

Simplifying the expression and evaluating at \(x=0\):

1. \(\frac{dy}{dx} \bigg|_{x=0} = \frac{1}{2}\)
2. Now find \(\frac{dz}{dx}\): 
   \(z = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)\)

Using the chain rule and quotient rule:

1. \(\frac{dz}{dx} = \frac{1}{1 + \left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)^2} \cdot \frac{d}{dx}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)\)

Find the derivative of the quotient:

1. \(\frac{d}{dx}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)\)

Using chain and quotient rules, and evaluating at \(x=0\):

1. \(\frac{dz}{dx} \bigg|_{x=0} = 2\)
2. Therefore, compute \(\frac{dy}{dz}\) at \(x=0\):

Thus, the correct answer is \(\frac{1}{4}\).