Find the derivative of `(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 \( y = \frac{u}{v} \), then the derivative \( y' \) is given by:

\[ y' = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = 1 - \cos x \) and \( v = 1 + \cos x \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = 1 - \cos x \)
- \( v = 1 + \cos x \)

### Step 2: Differentiate \( u \) and \( v \)
Now we need to find the derivatives of \( u \) and \( v \):
- The derivative of \( u \):
\[ \frac{du}{dx} = 0 - (-\sin x) = \sin x \]

- The derivative of \( v \):
\[ \frac{dv}{dx} = 0 + (-\sin x) = -\sin x \]

### 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 simplify the numerator:
\[ y' = \frac{(1 + \cos x)\sin x + (1 - \cos x)\sin x}{(1 + \cos x)^2} \]
\[ = \frac{\sin x (1 + \cos x + 1 - \cos x)}{(1 + \cos x)^2} \]
\[ = \frac{\sin x (2)}{(1 + \cos x)^2} \]
\[ = \frac{2 \sin x}{(1 + \cos x)^2} \]

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