= Solution
A dominated differentiation theorem is the following. Suppose $\phi(x,t)$ is measurable in $x$ and differentiable in $t$ on an open interval, is integrable for at least one parameter value, and its parameter derivative is dominated in absolute value by an <integrable function> independent of $t$. Then its integral is differentiable and
$$
\frac d{dt}\int\phi(x,t)\,dx=\int\partial_t\phi(x,t)\,dx.
$$
Local versions require the domination only on a neighbourhood of each parameter. The proof bounds the difference quotients by the dominating function, using the <fundamental theorem of calculus> or the mean-value theorem, then applies dominated convergence. Apply the argument to real and imaginary parts for a complex integrand.
For fixed $t$, $f(x)e^{itx}$ is measurable and has absolute value $|f(x)|$, so is integrable. Its derivative with respect to $t$ is $ixf(x)e^{itx}$, whose absolute value is the assumed <integrable function> $|xf(x)|$. Hence
$$
\boxed{\widehat f'(t)=\int_{\mathbb R}ixf(x)e^{itx}\,dx}.
$$
Dominated convergence also shows this derivative is continuous, though only differentiability is required.
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