Expression for Frequency-shifted Gaussian pulse
Well the expression for the $\frac {d^n}{dt^n} p(t) $ is $V_n(t) p(t)$, where $V_n(t)$ is a polynomial of degree $n$, and $V_n(t)$ may be determined recursivelyy
$$  V_{n+1}(t)e^{-t^2/(2\tau^2)} = \frac{d}{dt}\left(V_{n}(t)e^{-t^2/(2\tau^2)}\right) $$
Solving for $V_{n+2}(t)$ we have
$$ V_{n+1} = \frac{d}{dt} V_{n}(t) - \frac{t}{2\tau^2} V_n(t)$$
In order to link the two forms of the equation we need
$$\begin{eqnarray}
 g(t) &amp;=&amp;  p(t)cos(2\pi f_{c}t) \\
      &amp;=&amp; V_{TX}exp(-\frac{t^{2}}{2\tau^2})cos(2\pi f_{c}t) \\
      &amp;=&amp; \frac{d^{n}}{dt^{n}} \left( V_{TX }exp(-\frac{t^{2}}{2\tau^2}) \right) \\
      &amp;=&amp; p(t)V_n(t)
\end{eqnarray}$$
Thus $cos(2\pi f_c t) = V_n(t)$, the right hand side is a polynomial. As $n \to \infty$, the are the Hermite polynomials have some asymptotic expansions in terms of cossines. So the two may assume approximate values for certain values of $f_c$ and $\tau$.
Numerically evaluating the derivaties we see some resemblance with gausiand windowed cosine.
function hermite;
  tau=pi;
  t = tau * linspace(-5, 5, 1000);
  for order = 4:2:20
    subplot(3,3,order/2-1)
    a = -1/(2*tau^2);
    P = V_coefs(order, a);
    plot(t, polyval(P(end:-1:1), t).*exp(a*t.^2));
    title(num2str(order, 'n=%d'))
  end
  
end

function coefs = V_coefs(order, a)
  coefs = 1;
  for ii = 1:order
    coefs = [(1:(ii-1)).*coefs(2:end), 0, 0] + [0, coefs .* (2 * a)];
  end
end