etermine the following derivatives:
(i) d
dx
(∫ x
2
t sin t dt
)

To find the derivative of the given expression, let's first rewrite it clearly:

\[
\text{Let } F(x) = \frac{d}{dx} \left(\int_{2}^{x} t \sin t \, dt \right)
\]

To solve this, we can apply the **Leibniz rule** for differentiation under the integral sign, which is a special case of the Fundamental Theorem of Calculus. The Leibniz rule states:

\[
\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t, x) \, dt \right) = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) \, dt
\]

Here, the integral is from a constant \(a = 2\) to \(b(x) = x\). Therefore, the derivative simplifies as:

\[
F(x) = \frac{d}{dx} \left( \int_{2}^{x} t \sin t \, dt \right) = t \sin t \bigg|_{t=x} \cdot \frac{d}{dx}(x) - t \sin t \bigg|_{t=2} \cdot \frac{d}{dx}(2)
\]

Since \(\frac{d}{dx}(x) = 1\) and \(\frac{d}{dx}(2) = 0\), the derivative is:

\[
F(x) = x \sin x - 0 = x \sin x
\]

Thus, the derivative is:

\[
\boxed{x \sin x}
\]

---

Would you like further details or have any questions about the solution?

Here are some related questions you might consider:
1. What is the derivative of \( \int_{0}^{x} e^{t^2} \, dt \)?
2. How does the Leibniz rule apply when the limits of integration are both functions of \(x\)?
3. Can you compute the derivative \( \frac{d}{dx} \left(\int_{0}^{x^2} \cos(t) \, dt \right) \)?
4. What if the integrand involved \(x\) explicitly, like \( \int_{1}^{x} x^2 \sin t \, dt \)?
5. How would you differentiate \( \int_{a(x)}^{b(x)} f(t) \, dt \) when both limits are functions of \(x\)?

**Tip:** When applying the Leibniz rule, always identify if the limits of integration are constants or functions of \(x\), as this affects how the rule is applied.

Explore how to find the derivative d/dx (∫ x^2 sin t dt) using the Leibniz rule. This problem involves integral calculus and differentiation concepts, including the application of the Fundamental Theorem of Calculus. Suitable for college-level students.

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