#### Answer

$2x^2sin(x^6)+\int_{2}^{x^2} {sin (t^3)} dt$

#### Work Step by Step

Let's take the derivative of both sides, use the product rule and let's apply the fundamental theorem of Calculus (substitute the limits, and multiply by their derivatives)! $\frac{dy}{dx}=x(sin(x^6)\cdot 2x)+1\cdot\int_{2}^{x^2} {sin (t^3)}dt=2x^2sin(x^6)+\int_{2}^{x^2} {sin (t^3)} dt$