If ` y = sqrt((1+ sin x)/(1+ cos x)), " then" (dy)/(dx) ` is equal to

Options

A. `- (1)/(sqrt(2)) "cosec"^(2) (x)/(2)`
B. ` - (1)/(2sqrt(2)) "cosec"^(2) ((x)/(2))`
C. `(1)/(2) ""sec^(2) (x)/(2)`
D. `- (1)/(2sqrt(2)) ""sin^(2) ((x)/(2))`

Explanation

To find the derivative of the function \( y = \sqrt{\frac{1 + \sin x}{1 + \cos x}} \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \sqrt{\frac{1 + \sin x}{1 + \cos x}} \]
This can be rewritten as:
\[ y = \left( \frac{1 + \sin x}{1 + \cos x} \right)^{1/2} \]

### Step 2: Use the chain rule
To differentiate \( y \), we will use the chain rule. The derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{1 + \sin x}{1 + \cos x} \right)^{-1/2} \cdot \frac{d}{dx} \left( \frac{1 + \sin x}{1 + \cos x} \right) \]

### Step 3: Differentiate the inner function using the quotient rule
Let \( u = 1 + \sin x \) and \( v = 1 + \cos x \). We need to differentiate \( \frac{u}{v} \):
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Where:
- \( \frac{du}{dx} = \cos x \)
- \( \frac{dv}{dx} = -\sin x \)

So, substituting these into the quotient rule gives:
\[ \frac{d}{dx} \left( \frac{1 + \sin x}{1 + \cos x} \right) = \frac{(1 + \cos x)(\cos x) - (1 + \sin x)(-\sin x)}{(1 + \cos x)^2} \]

### Step 4: Simplify the derivative
Now we simplify the numerator:
\[ (1 + \cos x) \cos x + (1 + \sin x) \sin x = \cos x + \cos^2 x + \sin x + \sin^2 x \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ = \cos x + 1 + \sin x \]
Thus, we have:
\[ \frac{d}{dx} \left( \frac{1 + \sin x}{1 + \cos x} \right) = \frac{\cos x + 1 + \sin x}{(1 + \cos x)^2} \]

### Step 5: Substitute back into the derivative of \( y \)
Now substituting back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{1 + \sin x}{1 + \cos x} \right)^{-1/2} \cdot \frac{\cos x + 1 + \sin x}{(1 + \cos x)^2} \]

### Step 6: Final expression
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{\cos x + 1 + \sin x}{(1 + \cos x)^2 \sqrt{\frac{1 + \sin x}{1 + \cos x}}} \]