If `y=(x+1)/(x-1),` then what is `(dy)/(dx)` equal to?

Options

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

Explanation

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \frac{x+1}{x-1} \), we will use the quotient rule. The quotient rule states that if you have a function in the form \( y = \frac{f(x)}{g(x)} \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]

where:
- \( f(x) = x + 1 \)
- \( g(x) = x - 1 \)

### Step 1: Identify \( f(x) \) and \( g(x) \)
Here, we have:
- \( f(x) = x + 1 \)
- \( g(x) = x - 1 \)

### Step 2: Find the derivatives \( f'(x) \) and \( g'(x) \)
Now, we calculate the derivatives:
- \( f'(x) = \frac{d}{dx}(x + 1) = 1 \)
- \( g'(x) = \frac{d}{dx}(x - 1) = 1 \)

### Step 3: Apply the quotient rule
Using the quotient rule:

\[ \frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]

Substituting \( f(x) \), \( g(x) \), \( f'(x) \), and \( g'(x) \):

\[ \frac{dy}{dx} = \frac{(1)(x - 1) - (x + 1)(1)}{(x - 1)^2} \]

### Step 4: Simplify the numerator
Now simplify the numerator:

\[ \frac{dy}{dx} = \frac{x - 1 - (x + 1)}{(x - 1)^2} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{x - 1 - x - 1}{(x - 1)^2} = \frac{-2}{(x - 1)^2} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{-2}{(x - 1)^2} \]

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