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

Recognize that the expression inside the inverse tangent resembles the identity: \[ \tan(2\theta(A) = \frac{2\tan\theta}{1 - \tan^2\theta} \] 

Let \( \theta = \tan^{-1}(x) \), then \( \tan(2\theta A)) = \frac{2x}{1 - x^2} \)

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

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

Answer 2

To solve the problem, we need to find the derivative of the function: 

$ y = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) $

1. Let:
$ u = \frac{2x}{1 - x^2} $

2. Derivative of $y$ using chain rule:
$ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} $

3. Calculate $u^2$:
$ u^2 = \left(\frac{2x}{1 - x^2}\right)^2 = \frac{4x^2}{(1 - x^2)^2} $

4. Calculate $\frac{du}{dx}$ using quotient rule:
$ u = \frac{2x}{1 - x^2} $
Numerator $N = 2x$, Denominator $D = 1 - x^2$

$ \frac{du}{dx} = \frac{D \cdot \frac{dN}{dx} - N \cdot \frac{dD}{dx}}{D^2} = \frac{(1 - x^2)(2) - (2x)(-2x)}{(1 - x^2)^2} $

$ = \frac{2(1 - x^2) + 4x^2}{(1 - x^2)^2} = \frac{2 + 2x^2}{(1 - x^2)^2} = \frac{2(1 + x^2)}{(1 - x^2)^2} $

5. Substitute into derivative formula:
$ \frac{dy}{dx} = \frac{1}{1 + \frac{4x^2}{(1 - x^2)^2}} \cdot \frac{2(1 + x^2)}{(1 - x^2)^2} = \frac{1}{\frac{(1 - x^2)^2 + 4x^2}{(1 - x^2)^2}} \cdot \frac{2(1 + x^2)}{(1 - x^2)^2} $

6. Simplify denominator:
$ (1 - x^2)^2 + 4x^2 = (1 - 2x^2 + x^4) + 4x^2 = 1 + 2x^2 + x^4 = (1 + x^2)^2 $

7. Simplify derivative expression:
$ \frac{dy}{dx} = \frac{(1 - x^2)^2}{(1 + x^2)^2} \times \frac{2(1 + x^2)}{(1 - x^2)^2} = \frac{2}{1 + x^2} $

Final Answer:
The derivative is: $ { \frac{2}{1 + x^2} } $ (Option A)