The task is to find the derivative of the function $f(x) = \sqrt{3x + 2}$ using the definition of a derivative, which involves finding the limit of the difference quotient.

The derivative of a function $f(x)$ using the limit definition is given by:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step-by-step process:

1. Substitute the function $f(x)$:

   We are given $f(x) = \sqrt{3x + 2}$. So, we need to compute the difference quotient:

   $\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{3(x+h) + 2} - \sqrt{3x + 2}}{h}$

2. Simplify the expression:

   This can be simplified by multiplying both the numerator and denominator by the conjugate of the numerator:

   $\frac{\sqrt{3(x+h) + 2} - \sqrt{3x + 2}}{h} \times \frac{\sqrt{3(x+h) + 2} + \sqrt{3x + 2}}{\sqrt{3(x+h) + 2} + \sqrt{3x + 2}}$

   The numerator becomes:

   $\left( \sqrt{3(x+h) + 2} \right)^2 - \left( \sqrt{3x + 2} \right)^2 = \left( 3(x+h) + 2 \right) - (3x + 2) = 3h$

   So the expression simplifies to:

   $\frac{3h}{h \left( \sqrt{3(x+h) + 2} + \sqrt{3x + 2} \right)}$

3. Simplify further:

   Cancel out the $h$'s in the numerator and denominator:

   $\frac{3}{\sqrt{3(x+h) + 2} + \sqrt{3x + 2}}$

4. Take the limit as $h \to 0$:

   As $h \to 0$, $\sqrt{3(x+h) + 2} \to \sqrt{3x + 2}$. Therefore, the limit of the expression becomes:

   $f'(x) = \frac{3}{2\sqrt{3x + 2}}$

So, the derivative of the function $f(x) = \sqrt{3x + 2}$ is:

$f'(x) = \frac{3}{2\sqrt{3x + 2}}$

Would you like further clarification or more details on this? Here are some related questions:

1. How can you apply the limit definition of the derivative to other functions like polynomials or trigonometric functions?
2. What is the role of conjugates in simplifying limits involving square roots?
3. How can you interpret the derivative $f'(x)$ in terms of the slope of the curve $f(x)$?
4. How does the chain rule apply if we were to differentiate $f(x) = \sqrt{3x + 2}$ without using the limit definition?
5. What other methods can be used to find derivatives beyond the limit definition, such as implicit differentiation?

Tip: When working with square roots and limits, multiplying by the conjugate is often a useful technique to eliminate radicals.