1. ## Find derivative (dF/dt)

2. Originally Posted by zaurm
Let G be an antiderivative of sin.

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

But by definition, G' is sin.

So $\frac{dF}{dx}=2x\sin(x^2)$