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The given function is \( g(t) = \ln(3 + t^2) \).

To find the derivative \( g'(t) \), we'll use the chain rule and the derivative of the natural logarithm function. The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot u'(t) \), where \( u = 3 + t^2 \).

### Steps:

1. Set \( u = 3 + t^2 \).
2. The derivative of \( u \) with respect to \( t \) is \( \frac{d}{dt}(3 + t^2) = 2t \).
3. Now, applying the chain rule:
   \[
   g'(t) = \frac{1}{3 + t^2} \cdot 2t
   \]
4. Simplify the result:
   \[
   g'(t) = \frac{2t}{3 + t^2}
   \]

Thus, the derivative of \( g(t) \) is:
\[
g'(t) = \frac{2t}{3 + t^2}
\]

Would you like more details on any part of this? Here are some additional related questions:

1. What is the chain rule, and how does it apply here?
2. Can you explain the derivative of a logarithmic function more generally?
3. How would you apply this result in a specific problem, like finding critical points?
4. How does this derivative behave as \( t \to \infty \)?
5. What happens to \( g'(t) \) when \( t = 0 \)?

### Tip:
When differentiating functions inside a logarithm, always remember to apply the chain rule correctly by multiplying by the derivative of the inner function.

Find the derivative of g(t) = ln(3 + t^2).

This detailed solution shows how to find the derivative of g(t) = ln(3 + t^2) using the chain rule. The function involves logarithmic differentiation and provides key insights into calculus concepts. Suitable for students in Grades 11-12 or early university.

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