Question: If $G(x) = \int_{x^2}^{e^x} (\sin t)^4 \, dt$, then
$G'(x) =$
1. **State the problem:** We want to find the derivative $G'(x)$ of the function $$G(x) = \int_{x^2}^{e^x} (\sin t)^4 \, dt.$$
2. **Recall the Leibniz rule for differentiation under the integral sign:** If $$F(x) = \int_{a(x)}^{b(x)} f(t) \, dt,$$ then $$F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x).$$
3. **Identify the components:** Here, $$a(x) = x^2, \quad b(x) = e^x, \quad f(t) = (\sin t)^4.$$
4. **Compute the derivatives of the limits:**
$$a'(x) = \frac{d}{dx} x^2 = 2x,$$
$$b'(x) = \frac{d}{dx} e^x = e^x.$$
5. **Apply the Leibniz rule:**
$$G'(x) = (\sin(e^x))^4 \cdot e^x - (\sin(x^2))^4 \cdot 2x.$$
6. **Final answer:**
$$\boxed{G'(x) = e^x (\sin(e^x))^4 - 2x (\sin(x^2))^4}.$$