If ` y= sin^(-1)"" [(1-x^2)//(1 +x^2)]`. Then ` (dy)/(dx)=`

Options

A. `-2 //(1+x^2)`
B. `2//(1+x^2)`
C. `1//(2 +x^2)`
D. `2//(2-x^2)`

Answer

To solve the problem, we need to differentiate the function \( y = \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \) with respect to \( x \).

### Step-by-Step Solution:

1. **Substitution**:
We can use the substitution \( x = \tan(\theta) \). Then, we have:
\[ \frac{1 - x^2}{1 + x^2} = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]
Using the identity \( 1 - \tan^2(\theta) = \cos(2\theta) \) and \( 1 + \tan^2(\theta) = \sec^2(\theta) \), we can rewrite this as:
\[ \frac{\cos(2\theta)}{\sec^2(\theta)} = \cos(2\theta) \cdot \cos^2(\theta) = \cos(2\theta) \]

2. **Rewrite \( y \)**:
Thus, we can express \( y \) as:
\[ y = \sin^{-1}(\cos(2\theta)) \]

3. **Using the identity**:
We know that:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
Therefore, we can rewrite \( y \):
\[ y = \frac{\pi}{2} - \cos^{-1}(\cos(2\theta)) = \frac{\pi}{2} - 2\theta \]

4. **Finding \( \theta \)**:
Since \( x = \tan(\theta) \), we have:
\[ \theta = \tan^{-1}(x) \]

5. **Substituting back**:
Substitute \( \theta \) back into the equation for \( y \):
\[ y = \frac{\pi}{2} - 2\tan^{-1}(x) \]

6. **Differentiate \( y \)**:
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 0 - 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
The derivative of \( \tan^{-1}(x) \) is:
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]
Therefore:
\[ \frac{dy}{dx} = -2 \cdot \frac{1}{1 + x^2} = -\frac{2}{1 + x^2} \]

### Final Result:
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{2}{1 + x^2} \]