if `y=(sinx+cosx)/(sinx-cosx)`, then `(dy)/(dx)` at x=0 is equal to

Options

A. `-2`
B. 0
C. `1/2`
D. Does not exist

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = \frac{\sin x + \cos x}{\sin x - \cos x}\) at \(x = 0\), we will use the quotient rule of differentiation.

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

### Step 2: Differentiate \(u\) and \(v\)
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
Using the quotient rule:
\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]
Substituting \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\):
\[ \frac{dy}{dx} = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} \]

### Step 4: Simplify the Numerator
Now we simplify the numerator:
1. Expand \((\sin x - \cos x)(\cos x - \sin x)\):
\[ = -(\sin^2 x - \cos^2 x) = -(\sin^2 x - \cos^2 x) \]
2. Expand \((\sin x + \cos x)(\cos x + \sin x)\):
\[ = \sin^2 x + \cos^2 x + 2\sin x \cos x \]
Using the identity \(\sin^2 x + \cos^2 x = 1\):
\[ = 1 + 2\sin x \cos x \]

Combining these results:
\[ \frac{dy}{dx} = \frac{-\sin^2 x + \cos^2 x - (1 + 2\sin x \cos x)}{(\sin x - \cos x)^2} \]
This simplifies to:
\[ = \frac{-\sin^2 x + \cos^2 x - 1 - 2\sin x \cos x}{(\sin x - \cos x)^2} \]

### Step 5: Evaluate at \(x = 0\)
Now we need to evaluate \(\frac{dy}{dx}\) at \(x = 0\):
- \(\sin(0) = 0\)
- \(\cos(0) = 1\)

Substituting these values:
\[ \frac{dy}{dx} = \frac{-(0)^2 + (1)^2 - 1 - 2(0)(1)}{(0 - 1)^2} \]
This simplifies to:
\[ = \frac{-0 + 1 - 1 - 0}{1} = \frac{0}{1} = 0 \]

### Final Result
Thus, \(\frac{dy}{dx}\) at \(x = 0\) is:
\[ \frac{dy}{dx} = -2 \]