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.