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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?

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    You lost a negative sign on your final answer. You should have $2\left(-\sin(\pi)+\sin(4\pi)\right)$. – 2017-02-16

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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 $F'(t)=f(t)$.

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    .... well this is pretty much exactly what OP wrote in his question .... – 2017-02-16
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    @Surb I just wanted to make sure that my answer is correct. – 2017-02-16
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    @Ant This is what I understood from your post. But then, it seems to me that the comment of Tim Thayer is much more relevant than this answer (I might be wrong though). – 2017-02-16