y = `sec^(- 1)((x+1)/(x-1))+sin^(- 1)((x-1)/(x+1))`, x > 0. Find dy/dx

Answer

To find the derivative \( \frac{dy}{dx} \) for the function

\[ y = \sec^{-1}\left(\frac{x+1}{x-1}\right) + \sin^{-1}\left(\frac{x-1}{x+1}\right) \]

where \( x > 0 \), we can follow these steps:

### Step 1: Rewrite the Inverse Functions
We know that the inverse secant function can be expressed in terms of the inverse cosine function:

\[ \sec^{-1}(u) = \cos^{-1}\left(\frac{1}{u}\right) \]

Thus, we can rewrite:

\[ y = \cos^{-1}\left(\frac{x-1}{x+1}\right) + \sin^{-1}\left(\frac{x-1}{x+1}\right) \]

### Step 2: Use the Identity
We can use the identity:

\[ \cos^{-1}(u) + \sin^{-1}(u) = \frac{\pi}{2} \]

for \( u = \frac{x-1}{x+1} \). Therefore, we have:

\[ y = \frac{\pi}{2} \]

### Step 3: Differentiate
Since \( y \) is a constant \( \frac{\pi}{2} \), its derivative with respect to \( x \) is:

\[ \frac{dy}{dx} = 0 \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = 0 \]

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