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

Options

A. `sec^(2)x`
B. `(1)/(2) "sec"^(2) (x)/(2)`
C. `(-1)/(2) "cosec"^(2) (x)/(2)`
D. none of these

Explanation

To find the derivative of the function \( y = \sqrt{\frac{\sec x - 1}{\sec x + 1}} \), we will follow these steps:

### Step 1: Rewrite the function
We start by rewriting \( \sec x \) in terms of \( \cos x \):
\[ \sec x = \frac{1}{\cos x} \]
Thus, we can rewrite \( y \) as:
\[ y = \sqrt{\frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1}} = \sqrt{\frac{\frac{1 - \cos x}{\cos x}}{\frac{1 + \cos x}{\cos x}}} = \sqrt{\frac{1 - \cos x}{1 + \cos x}} \]

### Step 2: Simplify the expression
Next, we simplify the expression under the square root:
\[ y = \sqrt{\frac{1 - \cos x}{1 + \cos x}} \]

### Step 3: Use trigonometric identities
We can use the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \) and \( 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \):
\[ y = \sqrt{\frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)}} = \sqrt{\frac{\sin^2\left(\frac{x}{2}\right)}{\cos^2\left(\frac{x}{2}\right)}} = \tan\left(\frac{x}{2}\right) \]

### Step 4: Differentiate the function
Now we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left(\tan\left(\frac{x}{2}\right)\right) \]
Using the chain rule, we have:
\[ \frac{dy}{dx} = \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} \]
Thus,
\[ \frac{dy}{dx} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right) \]

### Final Answer
Therefore, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right) \]

---