Computing the nth-derivative $\frac{d^{n}}{d\lambda^{n}}e^{\lambda x-\frac{\lambda^{2}}{2}t}$ According to wolfram-alpha, $\frac{d^{n}}{d\lambda^{n}}e^{\lambda x-\frac{\lambda^{2}}{2}t}=  \frac{(-i)^{n} (-t)^{\frac{n}{2}} }{2^{\frac{n}{2}} }e^{x \lambda-\frac{t \lambda^2}{2}} H_n(\frac{(x-t \lambda)}{\sqrt{2t}}) $
,where $H_{n}(y)=(-1)^{n}e^{y^{2}}\frac{d^{n}}{dy^{n}}e^{-y^{2}}$. 
So I am wondering what will $H_n(\frac{(x-t \lambda)}{\sqrt{2t}})$ equal to; what will $\frac{d^{n}}{dy^{n}}$ be?
Also, how did wolfram-alpha compute the above derivative?

Also, is there a different way to write $\frac{d^{n}}{d\lambda^{n}}e^{\lambda x-\frac{\lambda^{2}}{2}t}$?
1)Using Taylor expansion $\frac{d^{n}}{d\lambda^{n}}e^{\lambda x-\frac{\lambda^{2}}{2}t}=\sum_{k=1}^{\infty}\frac{d^{n}}{d\lambda^{n}}\frac{(\lambda x-\frac{\lambda^{2}}{2}t)^{k}}{k!}$
Can we compute, $\frac{d^{n}}{d\lambda^{n}}(\lambda x-\frac{\lambda^{2}}{2}t)^{k}$? Wolfram-alpha doesn't answer it for some reason.
Thanks
 A: By using 
\begin{align}
D^{n}[ f(x) \, g(x) ] = \sum_{k=0}^{n} \binom{n}{k} \, f^{k}(x) \, g^{n-k}(x)
\end{align}
then it is seen that 
\begin{align}
D^{n} [ e^{a x} \, e^{- b x^{2}/2} ] &= e^{ax} \, \sum_{k=0}^{n} \binom{n}{k} \, D^{k}(e^{-bx^{2}/2}) \, a^{n-k} \\
&= e^{ax- bx^{2}/2} \, \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} \, \left(\frac{b}{2}\right)^{k/2} \, a^{n-k} \, H_{n}\left( \sqrt{\frac{b}{2}} x \right).  \\
\end{align}
Now using the formula
\begin{align}
H_{n}(x+y) = \sum_{k=0}^{n} \binom{n}{k} \, H_{k}(x) \, (2y)^{n-k}
\end{align}
then it is seen that
\begin{align}
H_{n}\left( \sqrt{\frac{b}{2}} \, x - \frac{a}{\sqrt{2b}} \right) = (-1)^{n} \left(\frac{b}{2}\right)^{n/2} \, \sum_{k=0}^{n} \binom{n}{k} \,  \left(- \frac{2a}{\sqrt{2b}}\right)^{n-k} \, H_{n}\left( \sqrt{\frac{b}{2}} x \right).
\end{align}
Now
\begin{align}
D^{n} [ e^{a x} \, e^{- b x^{2}/2} ] &= (-1)^{n} \left(\frac{2}{b}\right)^{n/2}
e^{ax- bx^{2}/2} \, H_{n}\left( \sqrt{\frac{b}{2}} \, x - \frac{a}{\sqrt{2b}} \right) 
\end{align} 
A: What you are looking for is the generalization of the chain rule for higher-order derivatives, the Faà di Bruno formula. I'll write its combinatorial form here, but the Wikipedia article has other forms of it:
 $$
 \frac{d^n}{d\lambda^n} f(g(\lambda))=(f\circ g)^{(n)}(\lambda)=\sum_{\pi\in\Pi} f^{(\left|\pi\right|)}(g(\lambda))\cdot\prod_{B\in\pi}g^{(\left|B\right|)}(\lambda)
$$
where the sum runs over all set partitions of $\{1,\cdots,n\}$ and the product runs over the blocks of a given partition. Let's work out the second derivative of your function with this formula. 
The partitions of the set $\{1,2\}$ are simply $\{\{1\},\{2\}\}$ and $\{\{1,2\}\}$. $f$ is the exponential function and $g(\lambda)= \lambda x-\frac{\lambda^2}{2}t$. Moreover, since your $f$ is the exponential function, the $f^{(|\pi|)}(g(x))$ factors out of the summation to given an overall factor $e^{\lambda x-\frac{\lambda^2}{2}t}$. We have
 $$
 \begin{align}
 \frac{d^2}{d\lambda^2} &= \left[\frac{d^2}{d\lambda^2}g(\lambda)+\left(\frac{dg(\lambda)}{d\lambda}\right)^2\right]e^{\lambda x-\frac{\lambda^2}{2}t}\\
       &=\left[-t+(x-\lambda t)^2\right]e^{\lambda x-\frac{\lambda^2}{2}t}
 \end{align}
 $$
which is indeed what you obtain by evaluating this explicitly.
You can simplify the higher derivatives by noting that since your $g(\lambda)$ is a quadratic terms, partitions with blocks of cardinality $>2$ will vanish. 
You can compute $\frac{d^n}{d\lambda^n}\left(\lambda x-\frac{\lambda^2}{2}\right)^k$ in the same manner, with $f=y^k$.
A: Let
\begin{equation}
u=u(\lambda)=\lambda x-\frac{\lambda^2}2t.
\end{equation}
Then, by the Faa di Bruno formula and some properties of the partial Bell polynomials,
\begin{align}
\frac{\operatorname{d}^n}{\operatorname{d}\lambda^n}\operatorname{e}^{\lambda x-\frac{\lambda^2}2t}
&=\sum_{k=0}^n(\operatorname{e}^{u})^{(k)} B_{n,k}\bigl(u'(\lambda), u''(\lambda), u'''(\lambda),\dotsc, u^{(n-k+1)}(\lambda)\bigr)\\
&=\operatorname{e}^{u}\sum_{k=0}^n B_{n,k}(x-\lambda t, -t, 0,\dotsc, 0)\\
&=\operatorname{e}^{\lambda x-\frac{\lambda^2}2t}\sum_{k=0}^n (-t)^k B_{n,k}\biggl(\lambda-\frac{x}{t}, 1, 0,\dotsc, 0\biggr)\\
&=\operatorname{e}^{\lambda x-\frac{\lambda^2}2t}\sum_{k=0}^n (-t)^k \frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}\biggl(\lambda-\frac{x}{t}\biggr)^{2k-n}\\
&=\frac{n!}{2^n}\operatorname{e}^{\lambda x-\frac{\lambda^2}2t} t^{n} \sum_{k=0}^n (-1)^k \frac{2^k}{k!}\binom{k}{n-k} \frac{(t\lambda-x)^{2k-n}}{t^k},
\end{align}
where we used the formula
\begin{equation}\label{Bell-x-1-0-eq}
B_{n,k}(x,1,0,\dotsc,0)
=\frac{1}{2^{n-k}}\frac{n!}{k!}\binom{k}{n-k}x^{2k-n}.
\end{equation}
For details of the above concepts, notions, and notations, please refer to related texts in the following papers.

*

*Feng Qi and Bai-Ni Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterranean Journal of Mathematics 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.

*Feng Qi and Bai-Ni Guo, Some properties of the Hermite polynomials, Georgian Mathematical Journal 28 (2021), no. 6, 925--935; available online at https://doi.org/10.1515/gmj-2020-2088.

*Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.

