`y=tan^(-1)((sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2))) `

Explanation

To differentiate the given expression \( y = \tan^{-1} \left( \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right) \) with respect to \( x \), we will follow these steps:

### Step 1: Substitute \( x^2 = \cos(2\theta) \)
We start by substituting \( x^2 = \cos(2\theta) \). This is useful because it allows us to simplify the square roots in the expression.

### Step 2: Rewrite the expression
Using the identity \( \cos(2\theta) = 2\cos^2(\theta) - 1 \) and \( 1 - \cos(2\theta) = 2\sin^2(\theta) \), we can rewrite the expression inside the arctangent:
\[ y = \tan^{-1} \left( \frac{1 + \cos(2\theta) + \sqrt{1 - \cos(2\theta)}}{1 + \cos(2\theta) - \sqrt{1 - \cos(2\theta)}} \right) \]

### Step 3: Simplify the expression
After substituting and simplifying, we get:
\[ y = \tan^{-1} \left( \frac{2\cos^2(\theta) + \sqrt{2\sin^2(\theta)}}{2\cos^2(\theta) - \sqrt{2\sin^2(\theta)}} \right) \]
This can be further simplified to:
\[ y = \tan^{-1} \left( \frac{\sqrt{2}\cos(\theta) + \sqrt{2}\sin(\theta)}{\sqrt{2}\cos(\theta) - \sqrt{2}\sin(\theta)} \right) \]

### Step 4: Use the tangent addition formula
Using the tangent addition formula \( \tan\left(\frac{\pi}{4} + \theta\right) = \frac{\tan(\frac{\pi}{4}) + \tan(\theta)}{1 - \tan(\frac{\pi}{4})\tan(\theta)} \), we can express:
\[ y = \tan^{-1}(\tan(\frac{\pi}{4} + \theta)) \]
Thus,
\[ y = \frac{\pi}{4} + \theta \]

### Step 5: Relate \( \theta \) back to \( x \)
Since we have \( \theta = \frac{1}{2} \cos^{-1}(x^2) \), we can express \( y \) as:
\[ y = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2) \]

### Step 6: Differentiate with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 0 + \frac{1}{2} \cdot \frac{d}{dx} \left( \cos^{-1}(x^2) \right) \]
Using the derivative of \( \cos^{-1}(u) \):
\[ \frac{d}{dx} \left( \cos^{-1}(u) \right) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]
where \( u = x^2 \) and \( \frac{du}{dx} = 2x \).

Thus,
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \left( -\frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x \right) = -\frac{x}{\sqrt{1 - x^4}} \]

### Final Answer
The derivative is:
\[ \frac{dy}{dx} = -\frac{x}{\sqrt{1 - x^4}} \]