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

Answer

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

### Step 1: Rewrite the Expression
Let \( y = \sec^{-1}\left(\frac{x-1}{x+1}\right) + \sin^{-1}\left(\frac{x+1}{x-1}\right) \).

### Step 2: Use the Identity for Inverse Functions
We know that:
\[ \sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2} \]
In our case, we can rewrite \( \sec^{-1}(u) \) in terms of \( \cos^{-1}(u) \):
\[ \sec^{-1}(u) = \cos^{-1}\left(\frac{1}{u}\right) \]
Thus, we can express \( y \) as:
\[ y = \cos^{-1}\left(\frac{x+1}{x-1}\right) + \sin^{-1}\left(\frac{x+1}{x-1}\right) \]

### Step 3: Combine the Terms
Since both terms inside the inverse functions are the same, we can apply the identity:
\[ y = \frac{\pi}{2} \]
This is because the argument \( \frac{x+1}{x-1} \) is the same for both functions.

### Step 4: Differentiate the Expression
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 0 \]
This is because \( y \) is a constant (\( \frac{\pi}{2} \)).

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

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