To find the derivative of \( F(t) = \ln(3t^2 - 4t + 12) \), we will use the chain rule and the derivative of the natural logarithm function.

The derivative of \( \ln(u(t)) \), where \( u(t) \) is a function of \( t \), is given by:
\[
\frac{d}{dt} \ln(u(t)) = \frac{1}{u(t)} \cdot u'(t)
\]

Here, \( u(t) = 3t^2 - 4t + 12 \).

### Step 1: Find the derivative of the inner function \( u(t) \):
\[
u'(t) = \frac{d}{dt} (3t^2 - 4t + 12) = 6t - 4
\]

### Step 2: Apply the chain rule:
\[
F'(t) = \frac{1}{3t^2 - 4t + 12} \cdot (6t - 4)
\]

### Final result:
\[
F'(t) = \frac{6t - 4}{3t^2 - 4t + 12}
\]

This is the derivative of the given function \( F(t) \).

Would you like further clarification or a deeper explanation?

Here are 5 related questions:
1. How do you apply the product rule and quotient rule together with chain rule?
2. What is the significance of natural logarithm in calculus?
3. How would the derivative change if the logarithmic function were different?
4. What does the second derivative of \( F(t) \) look like?
5. How does this derivative relate to the rate of change of the original function?

**Tip:** When differentiating composite functions, always check for hidden inner functions to apply the chain rule.

Learn how to find the derivative of the function F(t) = ln(3t^2 - 4t + 12) using the chain rule. This solution covers key calculus concepts such as logarithmic differentiation and the chain rule. Suitable for advanced high school and college students.

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