To find the derivative of the integral $\frac{d}{dx} \left( \int_{x^2}^{x} t \sin(t) \, dt \right)$, we can use the Leibniz rule for differentiation under the integral sign, which is an extension of the Fundamental Theorem of Calculus.

Leibniz Rule

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

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

In our case:

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

Applying 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 $t = b(x) = x$ and $t = a(x) = x^2$:

   • $f(b(x)) = x \sin(x)$
   • $f(a(x)) = x^2 \sin(x^2)$

Substitute into the Leibniz Rule

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

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

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

Final Answer

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

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 to integrals with variable limits?
2. What is the Leibniz rule, and how is it derived?
3. How can this method be extended to more complex integrands?
4. Can we use the same approach for definite integrals with constants as bounds?
5. How would the result change if the integrand were $\cos(t)$ instead of $\sin(t)$?

Tip: Always check the bounds and the function within the integral when applying the Leibniz rule to ensure proper differentiation.