We are tasked with finding the derivative $f'(-3)$ using the alternative form of the derivative. The function is given as:

$f(x) = x^3 + 3x^2 + 6$ where $c = -3$.

The alternative form of the derivative is defined as:

$f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$

We need to apply this formula with $c = -3$.

1. Step 1: Calculate $f(c)$
   Since $c = -3$, we substitute into the function:

   $f(-3) = (-3)^3 + 3(-3)^2 + 6 = -27 + 27 + 6 = 6$

2. Step 2: Set up the derivative limit

   Now apply the formula for the derivative:

   $f'(-3) = \lim_{x \to -3} \frac{f(x) - f(-3)}{x + 3}$

   Substituting the expression for $f(x) = x^3 + 3x^2 + 6$ and $f(-3) = 6$:

   $f'(-3) = \lim_{x \to -3} \frac{(x^3 + 3x^2 + 6) - 6}{x + 3}$

   Simplifying:

   $f'(-3) = \lim_{x \to -3} \frac{x^3 + 3x^2}{x + 3}$

3. Step 3: Factor the numerator

   We can factor $x^3 + 3x^2$ to simplify the limit:

   $x^3 + 3x^2 = x^2(x + 3)$

   Thus, the expression becomes:

   $f'(-3) = \lim_{x \to -3} \frac{x^2(x + 3)}{x + 3}$

4. Step 4: Cancel common factors

   The $x + 3$ terms cancel out:

   $f'(-3) = \lim_{x \to -3} x^2$

5. Step 5: Evaluate the limit

   Now we substitute $x = -3$ into the remaining expression:

   $f'(-3) = (-3)^2 = 9$

Thus, the derivative at $x = -3$ is:

$f'(-3) = 9$

Would you like more details on any of the steps? Here are a few related questions:

1. How do you use the first principles to find the derivative in general?
2. What is the geometric interpretation of the derivative?
3. Can the derivative of all functions be found using the alternative form?
4. How do you handle discontinuities when using limits to find derivatives?
5. What are some common mistakes when applying the alternative form of the derivative?

Tip: Factoring is often crucial when simplifying limits involving polynomials. Always check for common factors in the numerator and denominator.