To solve this, we need to find the derivative of the given function:

$t(x) = \ln(4x^3 + 4x^2 + 6x + 4)$

Step-by-step solution:

1. Use the chain rule:

The derivative of $\ln(f(x))$ is:

$\frac{d}{dx}[\ln(f(x))] = \frac{f'(x)}{f(x)}$

Here, $f(x) = 4x^3 + 4x^2 + 6x + 4$.

2. Find $f'(x)$:

Now we compute the derivative of $f(x)$.

$f'(x) = \frac{d}{dx} \left( 4x^3 + 4x^2 + 6x + 4 \right)$ Using power rule: $f'(x) = 12x^2 + 8x + 6$

3. Apply the chain rule:

Now apply the chain rule to get the derivative of $t(x)$:

$t'(x) = \frac{f'(x)}{f(x)} = \frac{12x^2 + 8x + 6}{4x^3 + 4x^2 + 6x + 4}$

Thus, the first derivative of $t(x)$ is:

$t'(x) = \frac{12x^2 + 8x + 6}{4x^3 + 4x^2 + 6x + 4}$

Correct Answer:

The correct answer matches the second option from the choices provided:

$\boxed{\frac{12x^2 + 8x + 6}{4x^3 + 4x^2 + 6x + 4}}$

Would you like any further details or clarifications on this solution?

Here are five related questions to expand on this:

1. How do we apply the chain rule in more complex logarithmic functions?
2. What are some common mistakes to avoid when differentiating logarithmic functions?
3. How can we simplify logarithmic derivatives further using algebraic techniques?
4. Can this logarithmic derivative be applied in any real-world problem contexts?
5. How does the derivative of natural logarithms differ from other types of logarithms?

Tip: When differentiating logarithmic functions, always ensure that the inside function's derivative is computed accurately before applying the quotient rule.