If y `= (cos x - sin x)/( cos x + sin x ) , "then " (dy)/(dx)`

Options

A. `x^(2)`
B. `y^(2)`
C. `1 + y^(2)`
D. `- (1 + y^(2))`

Answer

To find \(\frac{dy}{dx}\) for the function

\[ y = \frac{\cos x - \sin x}{\cos x + \sin x}, \]

we will use the quotient rule for differentiation. The quotient rule states that if you have a function in the form \(\frac{u}{v}\), then the derivative is given by:

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

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

### Step 1: Identify \(u\) and \(v\)

Let:
- \(u = \cos x - \sin x\)
- \(v = \cos x + \sin x\)

### Step 2: Differentiate \(u\) and \(v\)

Now we differentiate \(u\) and \(v\):

\[ \frac{du}{dx} = -\sin x - \cos x, \]
\[ \frac{dv}{dx} = -\sin x + \cos x. \]

### Step 3: Apply the Quotient Rule

Using the quotient rule:

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

### Step 4: Simplify the Numerator

Now we simplify the numerator:

1. Expand the first term:
\[ (\cos x + \sin x)(-\sin x - \cos x) = -\cos x \sin x - \cos^2 x - \sin^2 x - \sin x \cos x = -(\cos^2 x + \sin^2 x + 2\cos x \sin x). \]

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

2. Expand the second term:
\[ (\cos x - \sin x)(-\sin x + \cos x) = -\cos x \sin x + \cos^2 x + \sin^2 x - \sin x \cos x = \cos^2 x + \sin^2 x - 2\cos x \sin x. \]
Again, using \(\cos^2 x + \sin^2 x = 1\):
\[ 1 - 2\cos x \sin x. \]

### Step 5: Combine the Numerator

Now, combine the two results:
\[ -(1 + 2\cos x \sin x) - (1 - 2\cos x \sin x) = -1 - 2\cos x \sin x - 1 + 2\cos x \sin x = -2. \]

### Step 6: Write the Final Derivative

Thus, the derivative becomes:
\[ \frac{dy}{dx} = \frac{-2}{(\cos x + \sin x)^2}. \]

### Final Answer

The final answer is:
\[ \frac{dy}{dx} = \frac{-2}{(\cos x + \sin x)^2}. \]