= Solution
Apply the <Leibniz integral rule>: differentiate both moving endpoints and the integrand's explicit $x$ dependence. The <partial derivative> is $\partial_x\sin(xt^2)=t^2\cos(xt^2)$. Hence
$$
\boxed{\begin{aligned}
f'(x)={}&\beta e^{\beta x}\sin\left(xe^{2\beta x}\right)-\alpha e^{\alpha x}\sin\left(xe^{2\alpha x}\right)\\
&+\int_{e^{\alpha x}}^{e^{\beta x}}t^2\cos(xt^2)\,dt.
\end{aligned}}
$$
The signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If $\alpha=\beta$, the integral defining $f$ is identically zero and all displayed terms cancel, so
$$
\boxed{f'(x)=0\quad\text{when }\alpha=\beta.}
$$
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