If `y= (sin x + cos x)/( sin x- cos x)`, find `(dy)/(dx)`.

Options

A. `(2)/( ( sin x- cos x)^2)`
B. `(-2)/( (sin x - cosx)^2)`
C. `(-2)/( (sinx + cos x)^2)`
D. None of these

Answer

To find the derivative of the function \( y = \frac{\sin x + \cos x}{\sin x - \cos x} \), we will use the quotient rule for differentiation. The quotient rule states that if you have a function \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

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

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

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

### Step 2: Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
Now we differentiate \( u \) and \( v \):
- \( \frac{du}{dx} = \cos x - \sin x \)
- \( \frac{dv}{dx} = \cos x + \sin x \)

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

\[ \frac{dy}{dx} = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} \]

### Step 4: Simplify the numerator
Now we simplify the numerator:

1. Expand both products:
- \( (\sin x - \cos x)(\cos x - \sin x) = \sin x \cos x - \sin^2 x - \cos^2 x + \cos x \sin x = 2\sin x \cos x - (\sin^2 x + \cos^2 x) \)
- Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ 2\sin x \cos x - 1 \]

2. Expand the second product:
- \( (\sin x + \cos x)(\cos x + \sin x) = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x \)

3. Combine the results:
\[ \text{Numerator} = (2\sin x \cos x - 1) - (1 + 2\sin x \cos x) = 2\sin x \cos x - 1 - 1 - 2\sin x \cos x = -2 \]

### Step 5: Write the final derivative
Now substituting back into the derivative formula:

\[ \frac{dy}{dx} = \frac{-2}{(\sin x - \cos x)^2} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{-2}{(\sin x - \cos x)^2} \]