We are tasked with finding the derivative $f'(x)$ for the given function:

$f(x) = \ln(6x^3 + 3x^2 + 3x + 3)$

Step-by-Step Solution:

To differentiate $f(x) = \ln(g(x))$, we use the chain rule:

$f'(x) = \frac{1}{g(x)} \cdot g'(x)$

where $g(x) = 6x^3 + 3x^2 + 3x + 3$.

Step 1: Find $g'(x)$

Now, we differentiate $g(x)$:

$g(x) = 6x^3 + 3x^2 + 3x + 3$

Using the power rule:

$g'(x) = 18x^2 + 6x + 3$

Step 2: Apply the Chain Rule

Now apply the chain rule:

$f'(x) = \frac{1}{g(x)} \cdot g'(x)$

Substituting $g(x)$ and $g'(x)$:

$f'(x) = \frac{18x^2 + 6x + 3}{6x^3 + 3x^2 + 3x + 3}$

Final Answer:

This matches the first option in the list:

$\boxed{\frac{18x^2 + 6x + 3}{6x^3 + 3x^2 + 3x + 3}}$

Would you like further details or have any questions?

Here are 5 related questions to expand on this topic:

1. How do you apply the chain rule to more complex functions?
2. Can you explain why the derivative of $\ln(x)$ is $\frac{1}{x}$?
3. How would you approach finding higher-order derivatives of this function?
4. What is the role of logarithmic differentiation in simplifying certain problems?
5. How does the quotient rule compare with the chain rule in similar cases?

Tip: When differentiating logarithmic functions, always remember to apply the chain rule if the argument is more than a simple variable.