To find the derivative of the given integral expression, let's restate the problem:

We need to determine the derivative of:

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

We can apply the Leibniz rule for differentiation under the integral sign. This rule is useful for finding the derivative of an integral where the limits of integration are functions of $x$.

Leibniz Rule

For an integral of the form:

$F(x) = \int_{a(x)}^{b(x)} f(t) \, dt$

The derivative with respect to $x$ is:

$F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$

Here:

• $a(x) = x^2$ (the lower limit)
• $b(x) = x$ (the upper limit)
• $f(t) = t \sin(t)$

Steps to Apply the Leibniz Rule

1. Differentiate the upper limit $b(x) = x$ with respect to $x$:

   $b'(x) = 1$

2. Differentiate the lower limit $a(x) = x^2$ with respect to $x$:

   $a'(x) = 2x$

3. Evaluate the function $f(t) = t \sin(t)$ at the upper and lower limits:

   • At $t = b(x) = x$: $f(b(x)) = x \sin(x)$

   • At $t = a(x) = x^2$: $f(a(x)) = x^2 \sin(x^2)$

Substitute into the Leibniz Rule

Using the Leibniz rule:

$\frac{d}{dx} \left( \int_{x^2}^{x} t \sin(t) \, dt \right) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$

$= (x \sin(x) \cdot 1) - (x^2 \sin(x^2) \cdot 2x)$

$= x \sin(x) - 2x^3 \sin(x^2)$

Final Answer

$\frac{d}{dx} \left( \int_{x^2}^{x} t \sin(t) \, dt \right) = x \sin(x) - 2x^3 \sin(x^2)$

This expression represents the derivative of the given integral with variable limits.

Would you like more details or have any questions?

Here are some additional questions that might help expand your understanding:

1. How does the Fundamental Theorem of Calculus apply when the limits of integration are functions of $x$?
2. What is the significance of the Leibniz rule in differentiating integrals with variable limits?
3. How would the answer change if the upper and lower limits were reversed?
4. How can we extend this technique to integrals with more complex functions of $x$ as limits?
5. What would the derivative look like if the integrand was $t \cos(t)$ instead of $t \sin(t)$?

Tip: When dealing with integrals that have variable limits, always identify the limits and the function inside the integral before applying the Leibniz rule to differentiate.