Question

If \(y = tan^{-1}\{\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\}\), where \(|x| < 1\), then \(\frac{dy}{dx}\) is equal to

• A. \(\frac{-x}{\sqrt{1-x^4}}\)
• B. \(\frac{x}{\sqrt{1-x^4}}\)
• C. \(\frac{-2x}{\sqrt{1-x^4}}\)
• D. \(\frac{2x}{\sqrt{1-x^4}}\)

Hint:

Put x squared equal to cos 2 theta so the fraction becomes tan of (pi/4 - theta).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Expressions containing \(\sqrt{1 + x^2}\) and \(\sqrt{1 - x^2}\) can be simplified by substituting \(x^2 = \cos 2\theta\).

Step 2: Substitute.
Let \(x^2 = \cos 2\theta\). Then \(1 + x^2 = 2\cos^2\theta\) and \(1 - x^2 = 2\sin^2\theta\), so \(\sqrt{1 + x^2} = \sqrt{2}\cos\theta\) and \(\sqrt{1 - x^2} = \sqrt{2}\sin\theta\) (for \(\theta\) in the first quadrant).
\[ y = \tan^{-1}\left(\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}\right) = \tan^{-1}\left(\frac{1 - \tan\theta}{1 + \tan\theta}\right) = \tan^{-1}\tan\left(\frac{\pi}{4} - \theta\right) = \frac{\pi}{4} - \theta \]

Step 3: Go back to x.
Since \(2\theta = \cos^{-1}(x^2)\), we get \(y = \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}(x^2)\).

Step 4: Differentiate.
\[ \frac{dy}{dx} = -\frac{1}{2}\cdot\left(-\frac{1}{\sqrt{1 - x^4}}\right)\cdot 2x = \frac{x}{\sqrt{1 - x^4}} \]

Final Answer:
\(\dfrac{dy}{dx} = \dfrac{x}{\sqrt{1 - x^4}}\), option (B). \[ \boxed{\frac{x}{\sqrt{1-x^4}}} \]