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 the derivative \(\frac{dy}{dx}\) of the function \(y = \frac{\sin x + \cos x}{\sin x - \cos x}\) at \(x = 0\), we will follow these steps:

### Step 1: Differentiate the function using the Quotient Rule
The Quotient Rule states that if you have a function \(y = \frac{u}{v}\), then the derivative is given by:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, 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 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
Let's simplify the numerator:
1. Expand both products:
- First term: \((\sin x - \cos x)(\cos x - \sin x) = -(\sin^2 x - \cos^2 x)\)
- Second term: \((\sin x + \cos x)(\cos x + \sin x) = \sin^2 x + \cos^2 x\)

2. Combine these:
\[ -\sin^2 x + \cos^2 x - (\sin^2 x + \cos^2 x) = -2\sin^2 x \]

So, the numerator becomes:
\[ -2\sin^2 x \]

### Step 5: Write the derivative
Thus, we have:
\[ \frac{dy}{dx} = \frac{-2\sin^2 x}{(\sin x - \cos x)^2} \]

### Step 6: Evaluate at \(x = 0\)
Now we need to evaluate \(\frac{dy}{dx}\) at \(x = 0\):
1. Calculate \(\sin(0) = 0\) and \(\cos(0) = 1\):
- Numerator: \(-2\sin^2(0) = -2(0)^2 = 0\)
- Denominator: \((\sin(0) - \cos(0))^2 = (0 - 1)^2 = 1\)

Thus:
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{0}{1} = 0 \]

### Final Answer
The value of \(\frac{dy}{dx}\) at \(x = 0\) is \(0\).

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