What is the derivative of `sqrt((1+cosx)/(1-cosx))` ?

Options

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

Explanation

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

### Step 1: Rewrite the expression
We start with:
\[ y = \sqrt{\frac{1 + \cos x}{1 - \cos x}} \]

### Step 2: Use trigonometric identities
We can use the half-angle identities:
- \( 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \)
- \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \)

Substituting these into the expression for \( y \):
\[ y = \sqrt{\frac{2 \cos^2\left(\frac{x}{2}\right)}{2 \sin^2\left(\frac{x}{2}\right)}} \]
This simplifies to:
\[ y = \sqrt{\frac{\cos^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)}} = \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} = \cot\left(\frac{x}{2}\right) \]

### Step 3: Differentiate \( y \)
Now we need to find the derivative \( \frac{dy}{dx} \):
\[ y = \cot\left(\frac{x}{2}\right) \]
The derivative of \( \cot u \) is \( -\csc^2 u \), where \( u = \frac{x}{2} \). We also need to apply the chain rule:
\[ \frac{dy}{dx} = -\csc^2\left(\frac{x}{2}\right) \cdot \frac{d}{dx}\left(\frac{x}{2}\right) = -\csc^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} \]

### Step 4: Final result
Thus, the derivative is:
\[ \frac{dy}{dx} = -\frac{1}{2} \csc^2\left(\frac{x}{2}\right) \]