The Fundamental Theorem of Calculus, Part 1, states:

$\frac{d}{dx} \int_a^x f(t) \, dt = f(x),$

provided that $f(t)$ is continuous on the interval $[a, x]$.

Problem:

You are tasked with finding the derivative of:

$g(s) = \int_7^s (t - t^7)^8 \, dt.$

Solution:

1. Apply the Fundamental Theorem of Calculus (Part 1). Since the upper limit of the integral is $s$, the derivative of $g(s)$ is simply the integrand $f(t) = (t - t^7)^8$, evaluated at $t = s$:

   $g'(s) = \big( (t - t^7)^8 \big|_{t = s}.$

2. Substitute $t = s$ into the integrand:

   $g'(s) = (s - s^7)^8.$

Final Answer:

$g'(s) = (s - s^7)^8.$

Would you like further clarification or detailed steps?

---

Here are 5 questions to expand on this concept:

1. How does the Fundamental Theorem of Calculus Part 1 differ from Part 2?
2. What happens if the bounds of integration depend on $s$ in a more complex way?
3. Can you explain the conditions under which the Fundamental Theorem of Calculus applies?
4. How does this theorem relate to finding areas under curves?
5. What modifications would be required if the lower limit of the integral was $s$ instead?

Tip: When differentiating an integral with variable bounds, always check if the bounds depend on the variable directly or indirectly!