Differentiate from first principles:
6. `(2x+3)/( 3x+2)`

Answer

To differentiate the function \( f(x) = \frac{2x + 3}{3x + 2} \) from first principles, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \frac{2x + 3}{3x + 2} \).

### Step 2: Write the formula for the derivative using first principles
The derivative \( f'(x) \) is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 3: Calculate \( f(x + h) \)
We need to find \( f(x + h) \):
\[ f(x + h) = \frac{2(x + h) + 3}{3(x + h) + 2} = \frac{2x + 2h + 3}{3x + 3h + 2} \]

### Step 4: Substitute \( f(x + h) \) and \( f(x) \) into the derivative formula
Substituting into the derivative formula gives:
\[ f'(x) = \lim_{h \to 0} \frac{\frac{2x + 2h + 3}{3x + 3h + 2} - \frac{2x + 3}{3x + 2}}{h} \]

### Step 5: Simplify the expression
To simplify the expression, we need to combine the fractions in the numerator:
\[ = \lim_{h \to 0} \frac{(2x + 2h + 3)(3x + 2) - (2x + 3)(3x + 3h + 2)}{h(3x + 3h + 2)(3x + 2)} \]

### Step 6: Expand the numerator
Expanding both terms in the numerator:
1. \( (2x + 2h + 3)(3x + 2) = 6x^2 + 4x + 6xh + 6h + 6 \)
2. \( (2x + 3)(3x + 3h + 2) = 6x^2 + 6xh + 4x + 9h + 6 \)

Now substituting these back into the limit:
\[ = \lim_{h \to 0} \frac{(6x^2 + 4x + 6xh + 6h + 6) - (6x^2 + 6xh + 4x + 9h + 6)}{h(3x + 3h + 2)(3x + 2)} \]

### Step 7: Combine like terms
Combining like terms in the numerator:
\[ = \lim_{h \to 0} \frac{(6xh - 6xh) + (4x - 4x) + (6h - 9h)}{h(3x + 3h + 2)(3x + 2)} \]
This simplifies to:
\[ = \lim_{h \to 0} \frac{-3h}{h(3x + 3h + 2)(3x + 2)} \]

### Step 8: Cancel \( h \) and take the limit
Cancelling \( h \) from the numerator and denominator:
\[ = \lim_{h \to 0} \frac{-3}{(3x + 3h + 2)(3x + 2)} \]
As \( h \to 0 \), this becomes:
\[ = \frac{-3}{(3x + 2)(3x + 2)} = \frac{-3}{(3x + 2)^2} \]

### Final Answer
Thus, the derivative of the function \( f(x) = \frac{2x + 3}{3x + 2} \) is:
\[ f'(x) = \frac{-5}{(3x + 2)^2} \]