If `y = sec^(-1) ((x^(2) + 1)/(x^(2) -1)) " then " (dy)/(dx) =` ?

Options

A. `(-2)/((1 + x^(2)))`
B. `(2)/((1 + X^(2)))`
C. `(-1)/((1 - X^(2)))`
D. none of these

Explanation

To solve the problem where \( y = \sec^{-1} \left( \frac{x^2 + 1}{x^2 - 1} \right) \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the given function:
\[ y = \sec^{-1} \left( \frac{x^2 + 1}{x^2 - 1} \right) \]
To simplify the differentiation, we can use the identity for secant and cotangent. We know that:
\[ \sec^{-1}(z) = \theta \implies z = \sec(\theta) \]
Thus, we can set:
\[ \frac{x^2 + 1}{x^2 - 1} = \sec(\theta) \]

### Step 2: Use trigonometric identities
From the relationship \( \sec(\theta) = \frac{1}{\cos(\theta)} \), we can express \( \theta \) in terms of \( x \):
\[ \sec(\theta) = \frac{1}{\cos(\theta)} \implies \cos(\theta) = \frac{x^2 - 1}{x^2 + 1} \]
Now, we can use the identity \( \cos(2\theta) = 1 - \tan^2(\theta) \) to relate it back to \( x \).

### Step 3: Let \( x = \cot(\theta) \)
To make the differentiation easier, we can let:
\[ x = \cot(\theta) \implies \theta = \cot^{-1}(x) \]
This gives us:
\[ y = \sec^{-1} \left( \frac{\cot^2(\theta) + 1}{\cot^2(\theta) - 1} \right) \]
Using the identity \( \cot^2(\theta) + 1 = \csc^2(\theta) \) and \( \cot^2(\theta) - 1 = \csc^2(\theta) - 2 \), we can simplify further.

### Step 4: Differentiate
Now, we can differentiate \( y \) with respect to \( x \):
\[ y = 2 \cot^{-1}(x) \]
Using the derivative of \( \cot^{-1}(x) \):
\[ \frac{dy}{dx} = -\frac{2}{1 + x^2} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{2}{1 + x^2} \]