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 perform the following steps:

1. Let \(y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) and \(z = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)\).
2. We need to find \(\frac{dy}{dz}\), using the chain rule: \(\frac{dy}{dz} = \frac{dy}{dx} \div \frac{dz}{dx}\).
3. First, calculate \(\frac{dy}{dx}\):
   • The expression for \(y\) is: \(y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\).
   • Let \(\frac{\sqrt{1+x^2}-1}{x} = u\), then \(y = \tan^{-1}(u)\).
   • By chain rule, \(\frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx}\).
   • For \(u = \frac{\sqrt{1+x^2}-1}{x}\), use quotient rule:
     • \(u = \frac{\sqrt{1+x^2}-1}{x} \Rightarrow u = \frac{v}{w}\) where \(v = \sqrt{1+x^2}-1\) and \(w = x\).
     • The derivative of \(v\) with respect to \(x\) is \(\frac{dv}{dx} = \frac{x}{\sqrt{1+x^2}}\), and \(\frac{dw}{dx} = 1\).
     • Thus, \(\frac{du}{dx} = \frac{w \cdot \frac{dv}{dx} - v \cdot \frac{dw}{dx}}{w^2} = \frac{x \cdot \frac{x}{\sqrt{1+x^2}} - (\sqrt{1+x^2}-1) \cdot 1}{x^2}\).
4. Similarly, calculate \(\frac{dz}{dx}\):
   • The expression for \(z\) is: \(\tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)\).
   • Let \(\frac{2x\sqrt{1-x^2}}{1-2x^2} = v\), then \(z = \tan^{-1}(v)\).
   • By chain rule, \(\frac{dz}{dx} = \frac{1}{1+v^2} \cdot \frac{dv}{dx}\).
   • Differentiating \(v\) = \frac{2x\sqrt{1-x^2}}{1-2x^2}\) using quotient rule:
     • \(v = \frac{2x\sqrt{1-x^2}}{u\)}\), whe\)
     • The derivative of the numerator is \(\frac{d}{dx} (2x\sqrt{1-x^2})\\)
     • Applying product rule to \(2x\sqrt{1-x^2}\): \(\frac{d}{dx}(2x) \cdot \sqrt{1-x^2} + 2x \cdot \frac{d}{dx}(\sqrt{1-x^2})\\)
5. Evaluate each derivative at \(x=0\).
6. Plug the values into \(\frac{dy}{dz}\) and find the ratio at \(x = 0\).

After calculating, the derivative evaluates to \(\frac{1}{4}\) at \(x = 0\).