Let G be an antiderivative of sin.
Then by definition of the integral, $\displaystyle F(x)=G(x^2)-G(1)$
If you differentiate, by applying the chain rule, you get :
$\displaystyle \frac{dF}{dx}=2xG'(x^2)$ (the derivative of G(1) is 0 because it's a constante)
But by definition, G' is sin.
So $\displaystyle \frac{dF}{dx}=2x\sin(x^2)$