Question

If $ y = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) $, then $ \frac{dy}{dx} $ is:

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

Hint:

Key Fact: Use inverse trigonometric identities to simplify before differentiating.

Answer 1

The Correct Option is A

Observe that the expression within the inverse tangent matches the identity: \[ \tan(2\theta(A) = \frac{2\tan\theta}{1 - \tan^2\theta} \]

If \( \theta = \tan^{-1}(x) \), it follows that \( \tan(2\theta A)) = \frac{2x}{1 - x^2} \)

Therefore, \[ y = \tan^{-1}\left( \frac{2x}{1 - x^2} \right) = \tan^{-1}(\tan(2\tan^{-1}x)) = 2\tan^{-1}(x) \]

Differentiating both sides yields: \[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}x) = 2 \cdot \frac{1}{1 + x^2} \]