`tan ^(-1)(sqrt((1-cos 3 x)/(1+cos 3x)))`

Answer

To differentiate the function \( y = \tan^{-1} \left( \sqrt{\frac{1 - \cos 3x}{1 + \cos 3x}} \right) \), we can follow these steps:

### Step 1: Simplify the Inner Function
We start with the expression inside the arctangent:

\[ \sqrt{\frac{1 - \cos 3x}{1 + \cos 3x}} \]

Using the trigonometric identities:
- \( 1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right) \)
- \( 1 + \cos \theta = 2 \cos^2 \left( \frac{\theta}{2} \right) \)

We can rewrite \( 1 - \cos 3x \) and \( 1 + \cos 3x \):

\[ 1 - \cos 3x = 2 \sin^2 \left( \frac{3x}{2} \right) \]
\[ 1 + \cos 3x = 2 \cos^2 \left( \frac{3x}{2} \right) \]

Substituting these into the square root gives:

\[ \sqrt{\frac{2 \sin^2 \left( \frac{3x}{2} \right)}{2 \cos^2 \left( \frac{3x}{2} \right)}} = \sqrt{\frac{\sin^2 \left( \frac{3x}{2} \right)}{\cos^2 \left( \frac{3x}{2} \right)}} = \tan \left( \frac{3x}{2} \right) \]

### Step 2: Rewrite the Function
Now we can rewrite \( y \):

\[ y = \tan^{-1} \left( \tan \left( \frac{3x}{2} \right) \right) \]

Since \( \tan^{-1} \tan \theta = \theta \) for \( \theta \) in the principal range of \( \tan^{-1} \), we have:

\[ y = \frac{3x}{2} \]

### Step 3: Differentiate the Function
Now we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{3x}{2} \right) = \frac{3}{2} \]

### Final Answer
Thus, the derivative of the function is:

\[ \frac{dy}{dx} = \frac{3}{2} \]