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The given problem asks us to use **part 1 of the Fundamental Theorem of Calculus** to find the derivative of the function:

\[
F(x) = \int_x^1 \sin(t^2) \, dt
\]

### Part 1 of the Fundamental Theorem of Calculus states:
If \( G(x) = \int_{a}^{x} f(t) \, dt \), then \( G'(x) = f(x) \), provided \( f \) is continuous. 

However, in this case, the upper limit of the integral is a constant and the lower limit is \( x \). So, we apply the extended version, where the derivative of an integral with respect to its lower bound is negative:

\[
\frac{d}{dx} \left( \int_{x}^{1} f(t) \, dt \right) = -f(x)
\]

### Solution:
Here, \( f(t) = \sin(t^2) \). Applying the theorem:

\[
F'(x) = - \sin(x^2)
\]

Thus, the derivative of \( F(x) \) is:

\[
F'(x) = -\sin(x^2)
\]

Would you like further explanation on this? Here are some related questions to deepen your understanding:

1. How does the Fundamental Theorem of Calculus connect integration and differentiation?
2. Why is the derivative of the integral with respect to the lower limit negative?
3. Can this method be applied if both the upper and lower limits are functions of \( x \)?
4. How would the answer change if the integrand was a different function, like \( \cos(t^2) \)?
5. What happens if the bounds of integration are reversed in an integral?

**Tip:** The Fundamental Theorem of Calculus can simplify many problems involving derivatives of integrals—especially if the integrand's limits depend on \( x \).

Use part of the Fundamental Theorem of Calculus to find the derivative of F(x) = ∫(from x to 1) sin(t^2) dt.

Learn how to find the derivative of F(x) = ∫(from x to 1) sin(t^2) dt using Part 1 of the Fundamental Theorem of Calculus. This solution covers key calculus concepts, including derivatives and integration, suitable for university-level students.

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