To solve the problem, we need to compute $F'(x)$, where $F(x) = \int_{4}^{x^2} \sqrt{t} \, dt$.

Steps:

1. Recall the Fundamental Theorem of Calculus and Chain Rule:
   $\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) b'(x) - f(a(x)) a'(x)$

2. Identify $a(x)$ and $b(x)$:
   Here, $a(x) = 4$ (a constant), and $b(x) = x^2$. The function $f(t) = \sqrt{t}$.

3. Apply the formula:
   Since $a(x) = 4$ is constant, $a'(x) = 0$, so only the upper limit contributes: $F'(x) = f(b(x)) \cdot b'(x)$

4. Substitute $b(x) = x^2$ and $b'(x) = 2x$: $F'(x) = \sqrt{b(x)} \cdot b'(x)$ $F'(x) = \sqrt{x^2} \cdot 2x$

5. Simplify:
   Since $\sqrt{x^2} = x$ (for $x > 0$), the derivative becomes: $F'(x) = x \cdot 2x = 2x^2$

Final Answer:

The derivative $F'(x) = 2x^2$, which corresponds to option (D).

Would you like additional clarification or a detailed explanation?

Here are 5 follow-up questions for practice:

1. What happens if the lower limit is a function instead of a constant?
2. How does the Fundamental Theorem of Calculus apply when both limits are functions?
3. What if the integral contains another term dependent on $x$ inside?
4. Can you compute $F'(x)$ if $F(x) = \int_{x}^{x^2} \sqrt{t} \, dt$?
5. What happens if the upper limit is a fractional power of $x$?

Tip: Always pay attention to whether the limits of integration are constants or functions of $x$—this determines how the derivative is computed.