= Solution
{c}
With $G$ a standard <normal random variable>,
$$
\mathbb E f(X(t,x))
=\frac1{\sqrt{2\pi}}\int_{\mathbb R}
f(a(t)x+\sqrt{V(t)}\,z)e^{-z^2/2}dz.
$$
For $f\in C_b^1$, in the usual sense that $f$ and its continuous <derivative> are bounded, the <mean value theorem> bounds the difference quotient by $a(t)\|f'\|_\infty$. <Dominated convergence theorem> therefore allows differentiation inside the <integral>:
$$
\boxed{\frac d{dx}\mathbb E f(X(t,x))=a(t)\mathbb E f'(X(t,x))
=e^{\int_0^t\mu(s)ds}\mathbb E f'(X(t,x)).}
$$
This also follows pathwise from $X(t,x+h)-X(t,x)=a(t)h$. If the textual description were read as assuming only bounded differentiability, some control of $f'$ is needed for the displayed right side; the stated $C_b^1$ hypothesis supplies it. Gaussian smoothing itself still differentiates a bounded <measurable> $f$, using the <derivative> of the <Normal distribution> instead.
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