The problem requires finding the derivative $F'(x)$ of the given function:

$F(x) = \int_{x^2}^{0} 3t \, dt$

Step-by-step solution:

1. Rewrite the integral with proper bounds: Reversing the bounds of integration changes the sign of the integral: $F(x) = -\int_{0}^{x^2} 3t \, dt$

2. Use the Fundamental Theorem of Calculus and Chain Rule: Let $g(x) = x^2$. Then, $F'(x) = - \frac{d}{dx} \left[ \int_{0}^{g(x)} 3t \, dt \right]$ Applying the chain rule: $F'(x) = -\left[ 3g(x) \cdot g'(x) \right]$

3. Calculate $g(x) = x^2$ and its derivative $g'(x)$: $g(x) = x^2 \quad \text{and} \quad g'(x) = 2x$

4. Substitute $g(x)$ and $g'(x)$ into the expression: $F'(x) = -\left[ 3(x^2) \cdot (2x) \right]$

5. Simplify: $F'(x) = -6x^3$

Final Answer:

$F'(x) = -6x^3$

Let me know if you want further explanation or details!

Here are 5 related questions:

1. What is the Fundamental Theorem of Calculus, and how is it applied here?
2. Why does reversing the bounds of an integral change its sign?
3. How does the chain rule apply in this problem?
4. What happens to $F(x)$ if $x = 0$?
5. Can we generalize this process for any $F(x) = \int_{g(x)}^{h(x)} f(t) \, dt$?

Tip: Always check the order of the integral bounds before differentiating—it may require flipping the sign.