I've got a question about the following multi-derivative $$\frac{\text{d}^m}{\text{d}a^m} \left(e^{-2\mu a}\Phi \left(\frac{a-\mu u}{\sqrt{u}}\right)\right),$$ where $m> 0$ is an integer , $\mu>0$, and $u>0$ are constants, $a>0$, $\Phi(x) $ is the standard normal CDF, and $\phi(x)$ is the standard normal PDF.
I tried to find $$\frac{\text{d}^m}{\text{d}x^m} \left(e^{bx}\Phi \left(x\right)\right),$$ firstly, but it seems be very difficult to get a recursive answer.
Any help or idea would be the most grateful.
I'm not sure whether the following derivative can be used, $$\frac{\text{d}^m}{\text{d}x^m}\Phi \left(x\right)=(-1)^{m-1}H_{m-1}(x)\phi(x),$$ where $H_m(x)$ is Hermite polynomial defined as $$H_m(x)=(-1)^m e^{\frac{x^2}{2}}\frac{\text{d}^m}{\text{d}x^m}\left(e^{-\frac{x^2}{2}}\right).$$
By using $ e^{bx}\varphi(x) = e^{\frac{b^2}{2}}\varphi(x-b) $, we have
\begin{equation} \begin{split} &\frac{\text{d}^m}{\text{d}x^m} (e^{bx}\varphi(x)) \\ =& \frac{\text{d}^m}{\text{d}x^m} (e^{\frac{b^2}{2}}\varphi(x-b))\\ =& e^{\frac{b^2}{2}} \frac{\text{d}^m}{\text{d}x^m} (\varphi(x-b))\\ =& e^{\frac{b^2}{2}} (-1)^m H_m (x-b) \varphi(x-b). \end{split} \end{equation}
Therefore,
\begin{equation} \begin{split} \frac{\text{d}^m}{\text{d}x^m} \left( e^{bx} \Phi(x) \right)=& b^m e^{bx} \Phi(x) \\ & + e^{\frac{b^2}{2}} \varphi(x-b) \Big[ b^{m-1} +(-1)^1 H_1 (x-b) b^{m-2} \\ & +, \dotsc, +(-1)^{m-2} H_{m-2} (x-b)b +(-1)^{m-1} H_{m-1} (x-b) \Big] , \end{split} \end{equation} where $H_m(x)$ is Hermite polynomial defined as above.