Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 48

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

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)! $\frac{dy}{dx}=x(sin(x^6)\cdot 2x)+1\cdot\int_{2}^{x^2} {sin (t^3)}dt=2x^2sin(x^6)+\int_{2}^{x^2} {sin (t^3)} dt$