`y = sin ^(-1)((1 - x^(2))/(1+ x^(2))) 0 lt x lt 1`

Explanation

To find the derivative of the function \( y = \sin^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) for \( 0 < x < 1 \), we can follow these steps:

### Step 1: Substitute \( x = \tan(\theta) \)

Let \( x = \tan(\theta) \). Then, we can express \( 1 - x^2 \) and \( 1 + x^2 \) in terms of \( \theta \):

\[
1 - x^2 = 1 - \tan^2(\theta) = \frac{\cos^2(\theta) - \sin^2(\theta)}{\cos^2(\theta)} = \frac{\cos(2\theta)}{\cos^2(\theta)}
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