Find the derivative of the following functions from first principle. `(x+1)/(x-1)`

Answer

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

### Step 1: Define the function and apply the first principle
The first principle of derivatives states that:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Given \( f(x) = \frac{x+1}{x-1} \), we need to calculate \( f(x+h) \).

### Step 2: Calculate \( f(x+h) \)
Substituting \( x+h \) into the function:
\[ f(x+h) = \frac{(x+h)+1}{(x+h)-1} = \frac{x + h + 1}{x + h - 1} \]

### Step 3: Substitute into the derivative formula
Now, we substitute \( f(x+h) \) and \( f(x) \) into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h} \]

### Step 4: Simplify the expression
To simplify the expression, we need a common denominator for the two fractions in the numerator:
\[ \frac{x+h+1}{x+h-1} - \frac{x+1}{x-1} = \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{(x+h-1)(x-1)} \]
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:
\[ f'(x) = \lim_{h \to 0} \frac{(x^2 + hx - h - 1) - (x^2 + hx - h + 1)}{h(x+h-1)(x-1)} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{-2}{h(x+h-1)(x-1)} \]

### Step 5: Cancel \( h \) and take the limit
As \( h \) approaches 0, the expression simplifies:
\[ f'(x) = \lim_{h \to 0} \frac{-2}{(x-1)(x-1)} = \frac{-2}{(x-1)^2} \]

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