`tan^(-1)((sin x)/(1+cos x))`

Answer

To solve the problem \( y = \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) \), we will simplify the expression inside the inverse tangent function and then differentiate the result.

### Step-by-Step Solution:

1. **Recognize the Identity**:
We know that:
\[ \tan\left(\frac{x}{2}\right) = \frac{\sin x}{1 + \cos x} \]
This identity will help us rewrite the expression inside the inverse tangent.

2. **Rewrite the Function**:
Using the identity, we can rewrite \( y \):
\[ y = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) \]
Since the tangent and the arctangent functions are inverses, this simplifies to:
\[ y = \frac{x}{2} + n\pi \quad (n \in \mathbb{Z}) \]
However, we can ignore the \( n\pi \) term for the principal value, so:
\[ y = \frac{x}{2} \]

3. **Differentiate the Function**:
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \]

### Final Answer:
Thus, the derivative of \( y = \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \]