If `y=(sinx+cosx)/(sinx-cosx)`, then `(dy)/(dx)" at "x=0` is

Options

A. `-2`
B. 0
C. `1/2`
D. `2/3`

Answer

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

### Step 1: Rewrite the function
We start with:
\[ y = \frac{\sin x + \cos x}{\sin x - \cos x} \]
To simplify the differentiation, we can divide the numerator and the denominator by \(\cos x\):
\[ y = \frac{\frac{\sin x}{\cos x} + 1}{\frac{\sin x}{\cos x} - 1} = \frac{\tan x + 1}{\tan x - 1} \]

### Step 2: Differentiate using the quotient rule
Now we differentiate \(y\) using the quotient rule, which states that if \(y = \frac{u}{v}\), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, let \(u = \tan x + 1\) and \(v = \tan x - 1\).

Calculating the derivatives:
- \(\frac{du}{dx} = \sec^2 x\)
- \(\frac{dv}{dx} = \sec^2 x\)

Thus, applying the quotient rule:
\[ \frac{dy}{dx} = \frac{(\tan x - 1)(\sec^2 x) - (\tan x + 1)(\sec^2 x)}{(\tan x - 1)^2} \]

### Step 3: Simplify the expression
Now, we simplify the numerator:
\[ \frac{dy}{dx} = \frac{(\tan x - 1)\sec^2 x - (\tan x + 1)\sec^2 x}{(\tan x - 1)^2} \]
\[ = \frac{\sec^2 x \left((\tan x - 1) - (\tan x + 1)\right)}{(\tan x - 1)^2} \]
\[ = \frac{\sec^2 x \left(\tan x - 1 - \tan x - 1\right)}{(\tan x - 1)^2} \]
\[ = \frac{\sec^2 x (-2)}{(\tan x - 1)^2} \]
\[ = \frac{-2\sec^2 x}{(\tan x - 1)^2} \]

### Step 4: Evaluate at \(x = 0\)
Now we need to evaluate \(\frac{dy}{dx}\) at \(x = 0\):
- \(\tan(0) = 0\)
- \(\sec^2(0) = 1\)

Substituting these values:
\[ \frac{dy}{dx} \bigg|_{x=0} = \frac{-2 \cdot 1}{(0 - 1)^2} = \frac{-2}{1} = -2 \]

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