Differentiate the following w.r.t. x :
`l_(n)(sqrt((1-cosx)/(1+cosx)))`

Answer

To differentiate the function \( y = \ln\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) \) with respect to \( x \), we can follow these steps:

### Step 1: Simplify the function
We start with:
\[ y = \ln\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) \]
Using the property of logarithms, we can rewrite this as:
\[ y = \frac{1}{2} \ln\left(\frac{1 - \cos x}{1 + \cos x}\right) \]

### Step 2: Differentiate using the chain rule
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\frac{1 - \cos x}{1 + \cos x}} \cdot \frac{d}{dx}\left(\frac{1 - \cos x}{1 + \cos x}\right) \]

### Step 3: Differentiate the fraction
To differentiate \( \frac{1 - \cos x}{1 + \cos x} \), we use the quotient rule:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
where \( u = 1 - \cos x \) and \( v = 1 + \cos x \).

Calculating \( u' \) and \( v' \):
- \( u' = \sin x \)
- \( v' = -\sin x \)

Now applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{1 - \cos x}{1 + \cos x}\right) = \frac{\sin x (1 + \cos x) - (1 - \cos x)(-\sin x)}{(1 + \cos x)^2} \]
This simplifies to:
\[ \frac{\sin x (1 + \cos x) + \sin x (1 - \cos x)}{(1 + \cos x)^2} = \frac{\sin x (2)}{(1 + \cos x)^2} = \frac{2 \sin x}{(1 + \cos x)^2} \]

### Step 4: Substitute back into the derivative
Now substituting this back into our derivative:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1 + \cos x}{1 - \cos x} \cdot \frac{2 \sin x}{(1 + \cos x)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\sin x}{(1 - \cos x)(1 + \cos x)} \]

### Step 5: Use the identity
Using the identity \( 1 - \cos^2 x = \sin^2 x \):
\[ (1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x \]
Thus:
\[ \frac{dy}{dx} = \frac{\sin x}{\sin^2 x} = \frac{1}{\sin x} = \csc x \]

### Final Result
The derivative of \( y = \ln\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \csc x \]

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