Find the derivative of the following :
`(1-cosx)/(1+cosx)`.

Answer

To find the derivative of the function \( y = \frac{1 - \cos x}{1 + \cos x} \), we will use the quotient rule. The quotient rule states that if \( y = \frac{a}{b} \), then the derivative \( y' \) is given by:

\[ y' = \frac{b \cdot \frac{da}{dx} - a \cdot \frac{db}{dx}}{b^2} \]

where \( a = 1 - \cos x \) and \( b = 1 + \cos x \).

### Step 1: Identify \( a \) and \( b \)
Let:
- \( a = 1 - \cos x \)
- \( b = 1 + \cos x \)

### Step 2: Differentiate \( a \) and \( b \)
Now, we need to find the derivatives \( \frac{da}{dx} \) and \( \frac{db}{dx} \).

1. **Differentiate \( a \)**:
\[ \frac{da}{dx} = \frac{d}{dx}(1 - \cos x) = 0 + \sin x = \sin x \]

2. **Differentiate \( b \)**:
\[ \frac{db}{dx} = \frac{d}{dx}(1 + \cos x) = 0 - \sin x = -\sin x \]

### Step 3: Apply the Quotient Rule
Now we can apply the quotient rule:

\[ y' = \frac{(1 + \cos x)(\sin x) - (1 - \cos x)(-\sin x)}{(1 + \cos x)^2} \]

### Step 4: Simplify the numerator
Now, let's simplify the numerator:

1. Expand the terms:
\[ y' = \frac{(1 + \cos x)\sin x + (1 - \cos x)\sin x}{(1 + \cos x)^2} \]

2. Combine like terms:
\[ y' = \frac{\sin x + \cos x \sin x + \sin x - \cos x \sin x}{(1 + \cos x)^2} \]
\[ y' = \frac{2\sin x}{(1 + \cos x)^2} \]

### Final Answer
Thus, the derivative of the function \( y = \frac{1 - \cos x}{1 + \cos x} \) is:

\[ y' = \frac{2\sin x}{(1 + \cos x)^2} \]