= Solution
{c}
We first prove the differentiation theorem. Fix $t\in(0,b)$ and take nonzero $h\to0$ with $t+h\in(0,b)$. The <measurable> difference quotients
$$
q_h(x)=\frac{f(x,t+h)-f(x,t)}h
$$
converge pointwise to $f_t(x,t)$. The <mean value theorem> and the assumed derivative bound give $|q_h(x)|\leq g(x)$. In particular $f_t(\cdot,t)$ is <measurable> as a limit of quotients along a sequence, and its bound by $g$ proves integrability. The <dominated convergence theorem>, in its parameter-limit form, says that pointwise convergence with a common integrable bound permits convergence of the integrals. Applying it yields
$$
\boxed{\frac d{dt}\int_0^\infty f(x,t)dx
=\lim_{h\to0}\int_0^\infty q_h(x)dx
=\int_0^\infty f_t(x,t)dx.}
$$
The parameter-limit statement follows from the ordinary sequential theorem by applying it to every sequence of nonzero $h$ tending to zero.
For the specific function, $|f(x,t)|\leq e^{-x^2}$, which is integrable by the allowed fact. It is differentiable in $t$, with
$$
f_t(x,t)=-2xe^{-x^2}\sin(2xt),\qquad
|f_t(x,t)|\leq2xe^{-x^2}.
$$
The last majorant has integral one on $[0,\infty)$, using the substitution $u=x^2$. All three hypotheses therefore hold. The indicator of the parameter domain is one at every point under consideration and does not affect this derivative.
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