Differentiate the function `(1)/(2 x - 3)` by First Principle of differentiation.

Answer

To differentiate the function \( f(x) = \frac{1}{2x - 3} \) using the First Principle of Differentiation, we will follow these steps:

### Step 1: Define the function
We start by defining the function:
\[ f(x) = \frac{1}{2x - 3} \]

### Step 2: Write the expression for \( f(x+h) \)
Next, we need to find \( f(x+h) \):
\[ f(x+h) = \frac{1}{2(x+h) - 3} = \frac{1}{2x + 2h - 3} \]

### Step 3: Apply the First Principle of Differentiation
The First Principle of Differentiation states that:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Substituting \( f(x+h) \) and \( f(x) \) into this formula gives:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{2x + 2h - 3} - \frac{1}{2x - 3}}{h} \]

### Step 4: Simplify the expression
To simplify the expression, we need a common denominator for the fractions in the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{(2x - 3) - (2x + 2h - 3)}{h(2x + 2h - 3)(2x - 3)} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{(2x - 3) - (2x + 2h - 3)}{h(2x + 2h - 3)(2x - 3)} = \lim_{h \to 0} \frac{-2h}{h(2x + 2h - 3)(2x - 3)} \]

### Step 5: Cancel \( h \) from the numerator and denominator
We can cancel \( h \) from the numerator and denominator:
\[ f'(x) = \lim_{h \to 0} \frac{-2}{(2x + 2h - 3)(2x - 3)} \]

### Step 6: Evaluate the limit as \( h \to 0 \)
Now, we can substitute \( h = 0 \):
\[ f'(x) = \frac{-2}{(2x - 3)(2x - 3)} = \frac{-2}{(2x - 3)^2} \]

### Conclusion
Thus, the derivative of the function \( f(x) = \frac{1}{2x - 3} \) is:
\[ f'(x) = \frac{-2}{(2x - 3)^2} \]