Answer
$y' = \frac{3(3x+ 2)}{(3x+2)^3+1}$
Work Step by Step
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $y = \int ^{3x+2}_{1}\frac {t}{1+t^3}dt$
Since the upper bound is a function we need to use the chain rule:
u = 3x+2
u' = 3
$y = \int ^{u}_{1}\frac {t}{1+t^3}dt$
Since $F(u) = \int^{u}_{0}f(t)dt$
$F'(u) = f(u)$
Thus,
$y' = \frac{u}{1+u^3}u'$
$y' = \frac{3x+ 2}{(3x+2)^3+1}3$
$y' = \frac{3(3x+ 2)}{(3x+2)^3+1}$
