If ` y = sin^(-1) ((1 -x^(2))/(1 +x^(2))),0 lt x lt 1 " then " (dy)/(dx) ` is equal to

Options

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

Explanation

To find the derivative \( \frac{dy}{dx} \) for the given function \( y = \sin^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \), we can follow these steps:

### Step 1: Substitute \( x \) with \( \tan(\theta) \)
Let \( x = \tan(\theta) \). Then, \( \theta = \tan^{-1}(x) \). The range of \( x \) from \( 0 \) to \( 1 \) corresponds to \( \theta \) ranging from \( 0 \) to \( \frac{\pi}{4} \).

### Step 2: Rewrite \( y \) in terms of \( \theta \)
Substituting \( x \) into the function, we have:
\[ y = \sin^{-1} \left( \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \right) \]
Using the identity \( \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \cos(2\theta) \), we can rewrite \( y \):
\[ y = \sin^{-1}(\cos(2\theta)) \]

### Step 3: Use the co-function identity
Using the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \), we can express \( y \) as:
\[ y = \frac{\pi}{2} - 2\theta \]

### Step 4: Differentiate \( y \) with respect to \( x \)
Now, we need to differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - 2\theta \right) \]
Since \( \frac{\pi}{2} \) is a constant, its derivative is \( 0 \). Thus:
\[ \frac{dy}{dx} = -2 \frac{d\theta}{dx} \]

### Step 5: Find \( \frac{d\theta}{dx} \)
Using the derivative of \( \theta = \tan^{-1}(x) \):
\[ \frac{d\theta}{dx} = \frac{1}{1 + x^2} \]
Substituting this back, we have:
\[ \frac{dy}{dx} = -2 \cdot \frac{1}{1 + x^2} \]

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