Radon transform of the function $h(x_1,\ldots,x_n) = x_1 g(x_1,\ldots,x_n)$, where $g$ is the density of multivariate Gaussian $N(\mu,\Sigma)$ Given an absolutely integrable function $f:\mathbb R^n \to \mathbb R$, let $R[f]$ be its Radon transform defined for every $(w,b) \in (\mathbb R^n \setminus \{0\}) \times \mathbb R$ by
$$
R[f](w,b) := \int_H f(x)ds(x) = \int_{\mathbb R^n}f(x)\delta(b-x^\top w)\,dx,
$$
where $\delta$ is the Dirac distribution and $ds$ is area element on the hyperplane
$$
H:=\{x \in \mathbb R^n \mid x^\top w = b\}.
$$
In particular, if $P$ is a probability distribution on $\mathbb R^n$ with density $f$ and $z \sim P$, then we can inteprete $R[f](w,b)$ as the density of the random variable $z^\top w$ evaluated at the point $b$.
Now, let $g:\mathbb R^n \to \mathbb R$ be the density of the multivariate Gaussian distribution $N(\mu,\Sigma)$ with mean $\mu \in \mathbb R^n$ and covariance matrix $\Sigma \in \mathbb R^{n \times n}$. Since $z^\top w \sim N(\mu^\top w,w^\top \Sigma w)$ for $z \sim N(\mu,\Sigma)$, it is clear by virtue of the previous remark that,
$$
R[g](w,b) = \varphi\left(\frac{b-\mu^\top w}{\|w\|_\Sigma}\right),
\tag{1}
$$
where $\varphi$ is the density of the standard Gaussian distribution $N(0,1)$, and $\|w\|_\Sigma := (w^\top \Sigma w)^{1/2}$.
Consider the function $h:\mathbb R^n \to \mathbb R$ defined by $h(x)=x_1 g(x)$ for every $x=(x_1,\ldots,x_n) \in \mathbb R^n$.

Question. In the spirit of (1), what is an analytic formula for $R[h](w,b)$ ?

In the special case where $d=1$ so that $\Sigma=\sigma>0$ is just a scalar, a simple computation gives
$$
R[h](w,b) = \int_{-\infty}^{+\infty} x\varphi(x)\delta(b-wx)\,dx = \frac{b}{w}\varphi(\frac{b-w\mu}{2\sigma w}) = \frac{b}{w}R[g](w,b).
$$

Update
The accepted answer shows that
$$
R[h](w,b) = (\alpha(b-\mu_2)+\mu_1)R[g](w,b),
$$
where $\mu_1 := e_1^\top \mu$, $e_1=(1,0,\dots,0) \in \mathbb R^n$, $\mu_2 := w^\top \mu$, $\alpha := \dfrac{e_1^\top \Sigma w}{w^\top \Sigma w}$.
Of course, this can be generalized by replacing $e_1$ with any unit-vector $v$.
 A: $\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}$In this "Gaussian" setting especially, it is convenient to approximate the delta function by the normal distribution $N(0,\ep^2)$ with $\ep\downarrow0$, so that
\begin{equation*}
    R[f](w,b)=\lim_{\ep\downarrow0}R_\ep[f](w,b), \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
    R_\ep[f](w,b):=\int_{\R^n}dx\,\vpi_\ep(w^\top x-b)f(x), 
\end{equation*}
\begin{equation*}
    \vpi_\ep(t):=\frac1\ep 
    \vpi\Big(\frac t\ep\Big),  
\end{equation*}
and $\vpi$ is the standard normal density.
Now, for $h(x)\equiv x_1 g(x)$ and $g$ the density of $N(\mu,\Si)$, we can write
\begin{equation*}
    R_\ep[h](w,b)
    =\int_{\R^n}dx\,g(x) x_1 \vpi_\ep(w^\top x-b)
    =E e_1^\top X\, \vpi_\ep(w^\top X-b), 
\end{equation*}
where $e_1:=[1,0,\dots,0]^\top\in\R^{n\times1}=\R^n$ and $X\sim N(\mu,\Si)$ .
Note that the joint distribution of $e_1^\top X$ and $w^\top X$ is bivariate normal with respective means
\begin{equation*}
\mu_1:=e_1^\top\mu\quad\text{and}\quad\mu_2:=w^\top\mu, \tag{2}\label{2}    
\end{equation*}
respective standard deviations
\begin{equation*}
\si_1:=\sqrt{e_1^\top\Si e_1} \quad\text{and}\quad \si_2:=\sqrt{w^\top\Si w}, \tag{3}\label{3}
\end{equation*}
and correlation
\begin{equation*}
    \rho:=\frac{e_1^\top\Si w}{\si_1\si_2}. \tag{4}\label{4}
\end{equation*}
So, straightforward calculations yield
\begin{equation*}
    R_\ep[h](w,b)
    =
    \frac{ \rho  \si _1 \si _2 (b-\mu _2)+\mu _1 (\si
   _2^2+\ep ^2)}{\sqrt{2 \pi } (\si _2^2+\ep ^2){}^{3/2}}\,
   \exp\Big\{-\frac{(b-\mu _2){}^2}{2 (\si _2^2+\ep ^2)}\Big\}.  
\end{equation*}
Finally, by \eqref{1},
\begin{equation*}
    R[h](w,b)
    =
    \frac{\rho\si_1(b-\mu_2)+\mu_1\si_2}{\sqrt{2\pi}\,\si _2^2}\,
   \exp\Big\{-\frac{(b-\mu_2)^2}{2 \si _2^2}\Big\},   
\end{equation*}
with $\mu_1,\mu_2,\si_1,\si_2,\rho$ given by \eqref{2}--\eqref{4}.
