Find the derivative of `((x+1)/(x-1))` using first principle of differentiation.

Answer

To find the derivative of the function \( f(x) = \frac{x+1}{x-1} \) using the first principle of differentiation, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \frac{x+1}{x-1} \).

### Step 2: Apply the first principle of differentiation
According to the first principle of differentiation, the derivative \( f'(x) \) is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 3: Calculate \( f(x+h) \)
We need to find \( f(x+h) \):
\[ f(x+h) = \frac{(x+h) + 1}{(x+h) - 1} = \frac{x + h + 1}{x + h - 1} \]

### Step 4: Substitute into the limit expression
Now substitute \( f(x+h) \) and \( f(x) \) into the limit:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h} \]

### Step 5: Find a common denominator
To simplify the expression, we need a common denominator:
\[ \frac{x+h+1}{x+h-1} - \frac{x+1}{x-1} = \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{(x+h-1)(x-1)} \]

### Step 6: Expand the numerators
Expanding both numerators:
1. \( (x+h+1)(x-1) = x^2 - x + hx - h + x - 1 = x^2 + hx - h - 1 \)
2. \( (x+1)(x+h-1) = x^2 + hx - x + 1 - h = x^2 + hx - h + 1 \)

Now substituting back:
\[ f'(x) = \lim_{h \to 0} \frac{(x^2 + hx - h - 1) - (x^2 + hx - h + 1)}{h(x+h-1)(x-1)} \]

### Step 7: Simplify the numerator
The numerator simplifies to:
\[ (x^2 + hx - h - 1) - (x^2 + hx - h + 1) = -2 \]
Thus, we have:
\[ f'(x) = \lim_{h \to 0} \frac{-2}{h(x+h-1)(x-1)} \]

### Step 8: Cancel \( h \)
Now, we can cancel \( h \):
\[ f'(x) = \lim_{h \to 0} \frac{-2}{(x+h-1)(x-1)} \]

### Step 9: Evaluate the limit
As \( h \to 0 \):
\[ f'(x) = \frac{-2}{(x-1)(x-1)} = \frac{-2}{(x-1)^2} \]

### Final Result
Thus, the derivative of \( f(x) = \frac{x+1}{x-1} \) is:
\[ f'(x) = \frac{-2}{(x-1)^2} \]