1. **Problem Statement:** Find the derivative of the function $$y = \int_{4}^{3x+7} \frac{t}{1+t^3} \, dt$$ using the first part of the Fundamental Theorem of Calculus. 2. **Recall the Fundamental Theorem of Calculus, Part 1:** If $$F(x) = \int_{a}^{g(x)} f(t) \, dt,$$ then $$F'(x) = f(g(x)) \cdot g'(x).$$ This means the derivative of an integral with a variable upper limit is the integrand evaluated at the upper limit times the derivative of the upper limit. 3. **Identify components:** Here, $$f(t) = \frac{t}{1+t^3},$$ the upper limit is $$g(x) = 3x + 7,$$ and the lower limit is a constant 4. 4. **Compute derivative of the upper limit:** $$g'(x) = \frac{d}{dx}(3x + 7) = 3.$$ 5. **Apply the theorem:** $$y' = f(g(x)) \cdot g'(x) = \frac{3x + 7}{1 + (3x + 7)^3} \cdot 3 = \frac{3(3x + 7)}{1 + (3x + 7)^3}.$$ 6. **Final answer:** $$\boxed{y' = \frac{3(3x + 7)}{1 + (3x + 7)^3}}.$$