Find `(dy)/(dx)" if y=sin"^(-1) ((sqrtx-1)/(sqrtx+1))+sec^(-1) ((sqrtx+1)/(sqrtx-1))`

Answer

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

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

we can follow these steps:

### Step 1: Rewrite the function
We start with the given function:

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

### Step 2: Use the identity for secant inverse
Recall that:

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

Thus, we can rewrite the second term:

\[ \sec^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right) = \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right). \]

### Step 3: Combine the terms
Now we can combine the two terms:

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

### Step 4: Use the identity for sine and cosine inverse
We know that:

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

Since both terms have the same argument \(\frac{\sqrt{x}-1}{\sqrt{x}+1}\), we can simplify:

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

### Step 5: Differentiate both sides
Now, we differentiate both sides with respect to \(x\):

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right). \]

Since \(\frac{\pi}{2}\) is a constant, its derivative is:

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

### Final Answer
Thus, the final result is:

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

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