## Calculus: Early Transcendentals 8th Edition

$$y' = \frac{3(3x+2)}{1 + (3x+2)^3}$$
$$y = \int\limits^{3x+2}_1{\frac{t}{1+t^3}}dt$$ Using FTC 1, substitute in the upper bound for t and multiply by the derivative of the upper bound. $$y' = \frac{3x+2}{1 + (3x + 2)^3} \times (3x+2)'$$ Simplify. $$y' = \frac{3x+2}{1 + (3x + 2)^3} \times 3$$ $$y' = \frac{3(3x+2)}{1 + (3x+2)^3}$$