The given equation is:

$y = \ln(3 + x^2)$

Let's compute the derivative $\frac{dy}{dx}$ of this function using the chain rule.

The derivative of $\ln(u)$ is $\frac{1}{u} \cdot \frac{du}{dx}$, where $u = 3 + x^2$. Now, let’s differentiate step by step:

1. Differentiate the logarithmic part $\ln(3 + x^2)$: $\frac{dy}{dx} = \frac{1}{3 + x^2} \cdot \frac{d}{dx}(3 + x^2)$

2. The derivative of $3 + x^2$ with respect to $x$ is $2x$.

Therefore, the derivative $\frac{dy}{dx}$ becomes: $\frac{dy}{dx} = \frac{2x}{3 + x^2}$

This is the result of differentiating the given function.

Would you like further details or have any questions?

Here are some related questions for exploration:

1. What is the second derivative of this function?
2. How can you apply this derivative to find critical points?
3. What is the domain of this function?
4. Can you plot the graph of this function and its derivative?
5. How does the function behave as $x \to \infty$?

Tip: Always apply the chain rule carefully when differentiating composite functions like logarithms.