To solve this problem, we need to compute the derivative of the integral:

$\frac{d}{dx} \int_{-2x}^{2x} t^2 \sin(t^2) \, dt.$

Step 1: Leibniz Rule

The derivative of a definite integral with variable limits is handled using the Leibniz rule: $\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x).$ Here, $f(t) = t^2 \sin(t^2)$, $a(x) = -2x$, and $b(x) = 2x$.

Step 2: Applying the Rule

First, calculate the contributions of the upper and lower limits:

• For $b(x) = 2x$: $f(b(x)) = (2x)^2 \sin((2x)^2) = 4x^2 \sin(4x^2), \quad b'(x) = 2.$

• For $a(x) = -2x$: $f(a(x)) = (-2x)^2 \sin((-2x)^2) = 4x^2 \sin(4x^2), \quad a'(x) = -2.$

Substitute these into the Leibniz rule: [ \frac{d}{dx} \int_{-2x}^{2x} t^2 \sin(t^2) , dt = \big[4x^2 \sin(4x^2) \cdot 2\big] - \big[4x^2 \sin(4x^2) \cdot (-2)\big]. ]

Step 3: Simplify

$\frac{d}{dx} \int_{-2x}^{2x} t^2 \sin(t^2) \, dt = 8x^2 \sin(4x^2) + 8x^2 \sin(4x^2).$

$\frac{d}{dx} \int_{-2x}^{2x} t^2 \sin(t^2) \, dt = 16x^2 \sin(4x^2).$

Final Answer:

The correct choice is: $\boxed{16x^2 \sin(4x^2)}.$

Do you want a detailed explanation for any step, or would you like to explore related questions? Here are some suggestions:

1. How does the Leibniz rule work for different limits of integration?
2. What happens if the integral had only one variable limit?
3. How do trigonometric functions affect integration and differentiation?
4. What are the real-world applications of integrals with variable limits?
5. Can this method be extended to more complex functions?

Tip: Always check the symmetry of the function when dealing