d/dx( ∫ 3x to x^3 (t^3 + 1)^10 dt )
I am not sure how to do this, except I know that you are suppose to use chain rule
$\displaystyle \frac{d}{dx} \int^{x^{3}}_{3x} (t^{3}+1)^{10} \ dt = \frac{d}{dx} \Big( F(x^{3}) - F(3x) \Big) $ where F is an antiderivative of $\displaystyle (t^3+1)^{10} $
$\displaystyle = F'(x^{3}) \ \frac{d}{dx} (x^{3}) - F'(3x) \ \frac{d}{dx} (3x) = \Big((x^{3})^3+1\Big)^{10} * 3x^{2} - \Big((3x)^{3}+1\Big)^{10} *3$
$\displaystyle = 3x^{2} (x^9+1)^{10} - 3(27x^{3}+1)^{10} $