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

Answer

To solve the problem of differentiating the function \( y = \sin^{-1}\left(\frac{x-1}{x+1}\right) + \sec^{-1}\left(\frac{x+1}{x-1}\right) \), we can follow these steps:

### Step 1: Rewrite the Function
We start with the given function:
\[ y = \sin^{-1}\left(\frac{x-1}{x+1}\right) + \sec^{-1}\left(\frac{x+1}{x-1}\right) \]

### Step 2: Use the Property of Inverse Trigonometric Functions
We know that:
\[ \sec^{-1}(x) = \cos^{-1}\left(\frac{1}{x}\right) \]
Thus, we can rewrite the second term:
\[ \sec^{-1}\left(\frac{x+1}{x-1}\right) = \cos^{-1}\left(\frac{x-1}{x+1}\right) \]
Now, we can express \( y \) as:
\[ y = \sin^{-1}\left(\frac{x-1}{x+1}\right) + \cos^{-1}\left(\frac{x-1}{x+1}\right) \]

### Step 3: Apply the Identity for Inverse Functions
Using the identity:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
we can simplify \( y \):
\[ y = \frac{\pi}{2} \]

### Step 4: Differentiate the Function
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 0 \]
since the derivative of a constant is zero.

### Final Answer
Thus, the derivative of the given function is:
\[ \frac{dy}{dx} = 0 \]