To solve the problem:

We are given the expression:

$\frac{d}{dx} \left[ \int_5^{x^3} \left( e^t - \sin(t) \right) dt \right].$

This involves differentiation of an integral with a variable limit, so we use the Leibniz rule for differentiation under the integral sign:

$\frac{d}{dx} \left[ \int_a^{g(x)} f(t) \, dt \right] = f(g(x)) \cdot g'(x).$

Here:

• $f(t) = e^t - \sin(t)$,
• The upper limit of the integral is $g(x) = x^3$,
• The lower limit is constant $a = 5$.

Step-by-step Solution:

1. Apply the Leibniz Rule:

$\frac{d}{dx} \left[ \int_5^{x^3} \left( e^t - \sin(t) \right) dt \right] = \left( e^{x^3} - \sin(x^3) \right) \cdot \frac{d}{dx}(x^3).$

2. Differentiate $x^3$:

$\frac{d}{dx}(x^3) = 3x^2.$

3. Combine the results:

$\frac{d}{dx} \left[ \int_5^{x^3} \left( e^t - \sin(t) \right) dt \right] = \left( e^{x^3} - \sin(x^3) \right) \cdot 3x^2.$

Final Answer:

$\boxed{3x^2 \left( e^{x^3} - \sin(x^3) \right)}.$

Would you like further clarification or details on this solution?

Related Questions:

1. What is the Leibniz rule and how is it applied in other cases?
2. How would the result change if the lower limit were not constant?
3. What happens if the function inside the integral involves both $x$ and $t$?
4. Can this concept be extended to integrals with both limits dependent on $x$?
5. How does this solution connect to the Fundamental Theorem of Calculus?

Tip:

When solving problems with variable limits of integration, always carefully differentiate any terms dependent on the upper or lower limit, using the chain rule where necessary!