If `y = tan^(-1) ((1 + x^(2))/(1 - x^(2))) " then " (dy)/(dx) =` ?

Options

A. `(2x)/((1 + x^(4)))`
B. `(-2x)/((1 +x^(4)))`
C. `(x)/((1 + x^(4)))`
D. none of these

Explanation

Given the function \( y = \tan^{-1} \left( \frac{1 + x^2}{1 - x^2} \right) \), we need to find \(\frac{dy}{dx}\).

### Step-by-Step Solution:

1. **Substitute \( x^2 = \tan(\theta) \):**
\[ \frac{1 + x^2}{1 - x^2} = \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \]
We know that:
\[ \frac{1 + \tan(\theta)}{1 - \tan(\theta)} = \tan\left(\frac{\pi}{4} + \theta\right) \]
Therefore,
\[ y = \tan^{-1} \left( \frac{1 + x^2}{1 - x^2} \right) = \tan^{-1} \left( \tan\left(\frac{\pi}{4} + \theta\right) \right) \]
Simplifying, we get:
\[ y = \frac{\pi}{4} + \theta \]

2. **Express \(\theta\) in terms of \(x\):**
Since \( x^2 = \tan(\theta) \),
\[ \theta = \tan^{-1}(x^2) \]
Therefore,
\[ y = \frac{\pi}{4} + \tan^{-1}(x^2) \]

3. **Differentiate \( y \) with respect to \( x \):**
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} + \tan^{-1}(x^2) \right) \]
The derivative of a constant \(\frac{\pi}{4}\) is 0, so:
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}(x^2) \right) \]

4. **Use the chain rule to differentiate \(\tan^{-1}(x^2)\):**
Let \( u = x^2 \), then:
\[ \frac{d}{dx} \left( \tan^{-1}(u) \right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
Since \( u = x^2 \), we have:
\[ \frac{du}{dx} = 2x \]
Therefore:
\[ \frac{d}{dx} \left( \tan^{-1}(x^2) \right) = \frac{1}{1 + (x^2)^2} \cdot 2x = \frac{2x}{1 + x^4} \]

5. **Combine the results:**
\[ \frac{dy}{dx} = \frac{2x}{1 + x^4} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{2x}{1 + x^4} \]