$$I(x)=\int_{-\exp(x)}^{x^2}\cos(xt^2) \mathrm dt,\qquad I: \mathbb{R} \rightarrow \mathbb{R}$$
I want to determine $I'(x)$ by hand.
I thought about using Leibniz integral rule (are there other ways?) and ended up with
$$I'(x)=\cos(x^5)2x-\cos(x(-\exp(x))^2)(-\exp(x)))+\int_{-\exp(x)}^{x^2}-\sin(xt^2)t^2dt$$
But this doesn't get me any further because I can't compute that last integral. I would appreciate any tips/solutions. Thanks for your help in advance!
Hint: For $\int \sin(xt^2)t^2dt$, we have $$\int \sin(xt^2)t^2dt\\ =\frac12\int\sin(xt^2)tdt^2\\ =\frac12\int\sin(xu)\sqrt udu\\ =\frac12(\frac{\cos(xu)\sqrt u}x-\frac12\int\frac{\cos(xu)}{x\sqrt u}du)$$ And $$\int \frac{\cos(y)}{\sqrt y}dy=\sqrt{2\pi}\operatorname{C}(\sqrt{\frac{2y}\pi })$$ (Where $\operatorname{C}$ denotes Fresnel C funtion)