`(d)/(dx)((cosx+sinx)/(cosx-sinx))=`

Options

A. `sec2xtan2x`
B. `2sec2x(sec2x+tan2x)`
C. `tan2x`
D. `sec2x(sec2x-tan2x)`

Answer

To solve the problem \(\frac{d}{dx}\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right)\), we will use the quotient rule of differentiation. Here are the steps:

### Step 1: Define the function
Let
\[ f(x) = \frac{\cos x + \sin x}{\cos x - \sin x} \]

### Step 2: Apply the quotient rule
The quotient rule states that if \(f(x) = \frac{u}{v}\), then
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
where \(u = \cos x + \sin x\) and \(v = \cos x - \sin x\).

### Step 3: Differentiate \(u\) and \(v\)
First, we find \(u'\) and \(v'\):
- For \(u = \cos x + \sin x\):
\[ u' = -\sin x + \cos x \]
- For \(v = \cos x - \sin x\):
\[ v' = -\sin x - \cos x \]

### Step 4: Substitute into the quotient rule
Now substitute \(u\), \(u'\), \(v\), and \(v'\) into the quotient rule:
\[ f'(x) = \frac{(-\sin x + \cos x)(\cos x - \sin x) - (\cos x + \sin x)(-\sin x - \cos x)}{(\cos x - \sin x)^2} \]

### Step 5: Simplify the numerator
Now we simplify the numerator:
1. Expand the first term:
\[ (-\sin x + \cos x)(\cos x - \sin x) = -\sin x \cos x + \sin^2 x + \cos^2 x - \cos x \sin x = 1 - 2\sin x \cos x \]
2. Expand the second term:
\[ (\cos x + \sin x)(-\sin x - \cos x) = -\cos x \sin x - \cos^2 x - \sin^2 x - \sin x \cos x = -1 - 2\sin x \cos x \]
3. Combine both:
\[ 1 - 2\sin x \cos x + 1 + 2\sin x \cos x = 2 \]

### Step 6: Write the final derivative
Now we have:
\[ f'(x) = \frac{2}{(\cos x - \sin x)^2} \]

### Final Answer
Thus, the derivative of the function is:
\[ \frac{d}{dx}\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) = \frac{2}{(\cos x - \sin x)^2} \]

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