`sec^(-1)((1+x^(2))/(1-x^(2)))`

Answer

To differentiate the function \( y = \sec^{-1}\left(\frac{1+x^2}{1-x^2}\right) \), we will follow these steps:

### Step 1: Rewrite the function
Let
\[ y = \sec^{-1}\left(\frac{1+x^2}{1-x^2}\right) \]

### Step 2: Substitute \( x \) with \( \tan \theta \)
To simplify the differentiation, we will use the substitution \( x = \tan \theta \). Then,
\[ y = \sec^{-1}\left(\frac{1+\tan^2 \theta}{1-\tan^2 \theta}\right) \]

### Step 3: Use the identity for secant
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \) and \( 1 - \tan^2 \theta = \cos 2\theta \), we can rewrite the expression:
\[ \frac{1+\tan^2 \theta}{1-\tan^2 \theta} = \frac{\sec^2 \theta}{\cos 2\theta} \]
This simplifies to:
\[ y = \sec^{-1}(\sec 2\theta) \]
Since \( \sec^{-1}(\sec x) = x \), we have:
\[ y = 2\theta \]

### Step 4: Express \( \theta \) in terms of \( x \)
From our substitution \( x = \tan \theta \), we can express \( \theta \) as:
\[ \theta = \tan^{-1}(x) \]
Thus, substituting back, we get:
\[ y = 2\tan^{-1}(x) \]

### Step 5: Differentiate \( y \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \]
Thus,
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \]

### Final Answer
The derivative of the function is:
\[ \frac{dy}{dx} = \frac{2}{1+x^2} \]

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