Find $g'(1)$ where $g(x) = \int_{x^2}^{2x}\, \sin(\pi u^2)\,du $ 
Find $g'(1)$, where $$g(x)=\int_{x^2}^{2x}\, \sin(\pi u^2)\,du, $$

I just want to make sure that my work is correct.
I started with setting $f(u)=\sin(\pi u^2)$ then I used some properties:
$(-F'(x^2)2x + 2F'(2x))$
therefore, 
\begin{align}
g'(-1) &= -2f(1) + 2f(2)\\
&= 2(-f(1)+f(2))\\
&= 2(\sin(\pi) + \sin(4\pi))
\end{align}
is it correct?

Hint
Let $f:u\mapsto \sin(\pi u^2 )$ and
$F:x\mapsto \int_0^x f(u)du$.
then
$$g(x)=F(2x)-F(x^2)$$ and
$$g'(x)=2F'(2x)-2xF'(x^2)$$
with $F'(t)=f(t)$.
