We are tasked with finding the derivative $f'(x)$ for the function $f(x) = \ln\left((x^2 + 4)^3\right)$.

Steps to solve:

1. Simplify the logarithmic function:
   Using the property of logarithms, $\ln(a^b) = b \ln(a)$, we can simplify: $f(x) = \ln\left((x^2 + 4)^3\right) = 3 \ln(x^2 + 4)$

2. Differentiate $f(x) = 3 \ln(x^2 + 4)$: To find the derivative, we apply the chain rule. $f'(x) = 3 \cdot \frac{d}{dx} \left[\ln(x^2 + 4)\right]$ The derivative of $\ln(u)$ with respect to $x$ is $\frac{1}{u} \cdot \frac{du}{dx}$, where $u = x^2 + 4$.

   First, calculate $\frac{d}{dx}(x^2 + 4)$: $\frac{d}{dx}(x^2 + 4) = 2x$

   Now, apply this to the chain rule: $f'(x) = 3 \cdot \frac{1}{x^2 + 4} \cdot 2x = \frac{6x}{x^2 + 4}$

Thus, the derivative is: $f'(x) = \frac{6x}{x^2 + 4}$

Let me know if you need more details, or if you'd like to explore further!

Here are 5 related questions to practice:

1. How would the derivative change if the power inside the logarithm was $5$ instead of $3$?
2. What is the second derivative $f''(x)$ of this function?
3. Can you generalize the result for $f(x) = \ln\left((x^2 + c)^n\right)$ for any constants $c$ and $n$?
4. What happens if you differentiate $f(x) = \ln\left((x^n + b)^k\right)$?
5. How would the derivative change if the function was $\ln(x^2 + 4)$ without the exponent $3$?

Tip: Always simplify logarithmic expressions when possible before differentiating to make the process easier.