# $n$-th derivative of $\exp\left(-\frac{\lambda(x-\mu)^2}{2\mu^2x}\right)$

Let $$\lambda$$ and $$\mu$$ be two positive real numbers and let denote $$f$$ the function defined as:

$$\forall x>0,~f(x):= \exp\left(-\frac{\lambda(x-\mu)^2}{2\mu^2x}\right).$$

I am struggling to find out an explicit expression of $$\partial_x^n f$$. According to this answer, it seems that:

$$\partial_x^n f(x) = f(x)\frac{P_{\lambda, \mu}(x)}{2^n \mu^{2n}x^{2n}}$$

where $$P_{\lambda, \mu}$$ is a polynomial function of $$x$$ depending on $$\lambda$$ and $$\mu$$.

Does anyone know how to deal with this derivative ?

Any hint, solution or reference will be highly appreciated!

Thank you.

• Of course $P_{\lambda, \mu}(x)$ also depends on $n$. For $n=3$, $$P_{\lambda, \mu}(x) =-8\,{\lambda}^{3}{x}^{6}+24\,{\lambda}^{3}{\mu}^{2}{x}^{4}+96\,{ \lambda}^{2}{\mu}^{4}{x}^{3}-8\,\lambda\, \left( 3\,{\lambda}^{2}{\mu} ^{4}-24\,{\mu}^{6} \right) {x}^{2}-96\,{\lambda}^{2}{\mu}^{6}x+8\,{ \lambda}^{3}{\mu}^{6}$$ Commented Apr 21 at 1:04
• Not sure how to use it but changing the variable to $t=\sqrt{\frac\lambda{2x}}\left(1-\frac x\mu\right)$ should presumably give expressions through Hermite polynomials in $t$ Commented Apr 21 at 4:07

$$\newcommand\la\lambda$$Let $$g(x):=-\frac\la{2\mu^2}\frac{(x-\mu)^2}x =-\frac\la{2\mu^2}(x-2\mu+\mu^2/x)$$ so that $$f(x)=e^{g(x)}.$$ So, by the Faà di Bruno formula, $$f^{(n)}(x)=f(x)\sum_{\pi\in\Pi_n}\prod_{B\in\pi}g^{(|B|)}(x),$$ where $$\Pi_n$$ is the set of all partitions of the set $$\{1,\dots,n\}$$, $$|B|$$ is the cardinality of $$B$$, and $$g^{(k)}(x)=-\frac\la{2\mu^2}\,1(k=1)-\frac\la2\,(-1)^k k!x^{-k-1}$$ for $$k=1,2,\dots$$.

For $$y_n:=f^{(n)}(x)/n!$$, Mathematica gives the recursion $$y(n+2) \left(-\lambda \mu ^2+4 \mu ^2 n x+\lambda x^2+8 \mu ^2 x\right)+2 \mu ^2 (n+3) x^2 y(n+3)+2 y(n+1) \left(\mu ^2+\mu ^2 n+\lambda x\right)+\lambda y(n)=0$$ for $$n\ge0$$, which implies that $$f^{(n)}(x)=f(x)n!\frac{P_n(x)}{2^n \mu^{2n}x^{2n}},$$ where, for each $$n$$, $$P_n(x)=P_{n;\la,\mu}(x)$$ is a polynomial such that $$P_0(x)=1,\quad P_1(x)=\lambda \left(\mu ^2-x^2\right),\quad P_2(x)=\frac{1}{2} \lambda \left(\lambda \mu ^4+\lambda x^4-2 \lambda \mu ^2 x^2-4 \mu ^4 x\right),$$ and the recurrence $$4 \lambda \mu ^4 x^4 P_n(x)+4 \mu ^2 x^2 P_{n+1}(x) \left(\mu ^2 (n+1)+\lambda x\right)+P_{n+2}(x) \left(-\lambda \mu ^2+4 \mu ^2 (n+2) x+\lambda x^2\right)+(n+3) P_{n+3}(x)=0$$ holds for $$n\ge0$$.

• Thank you @Iosif Pinelis! Commented Apr 21 at 15:24

The polynomials $$P$$ are related to the generalized Laguerre polynomials, see A066667. After some algebra, the $$n$$-th derivative of $$f(x)$$ can be written as \begin{align}\label{eq:1}\tag{1} \frac{\partial^n_x f(x)}{f(x)}= \left(-\frac{\lambda }{2\mu ^2}\right)^n + \sum_{k=1}^n \frac{n!}{(n-k)!} \, \frac{(-1)^k \mu ^2 }{k \, x^{k+1}} \left(-\frac{\lambda }{2\mu^2}\right)^{n-k+1} L_{k-1}^{(1)}\left(\frac{\lambda }{2 x}\right), \end{align} with the generalized Laguerre polynomials $$L^{(\alpha)}_n(x)$$.

• Thank you @Fred Hucht for your answer! Could you be a bit more explicit on the main steps of "some algebra" ? Thank you! Commented Apr 22 at 12:13
• I expanded the LHS of (1) for $n=3,\ldots,7$ using Mathematica's Apart[] (with the substitution $\lambda \mapsto 2 t x$). The resulting terms, e.g., $5040 - 15120 t + 12600 t^2 - 4200 t^3 + 630 t^4 - 42 t^5 + t^6$ for $n=7$, were looked up and identified as generalized Laguerre polynomials using oeis.org. Finally, the prefactors were adjusted. Commented Apr 22 at 13:00

$$\newcommand\la\lambda$$Here is a very simple expression for $$f^{(N)}(x)$$:

Note that $$f(x)=g(x)h(bx),$$ where $$g(x):=e^{a(x-2\mu)},\quad h(u):=e^{1/u},\quad a:=-\frac\la{2\mu^2},\quad b:=-\frac2\la.$$ So, by the Leibniz differentiation formula, $$f^{(N)}(x)=g(x)\sum_{n=0}^N\binom Nn a^{N-n} b^n h^{(n)}(bx).$$ In turn, it is easily checked by induction on $$n$$ that $$h^{(n)}(u)=h(u)\sum_{k=0}^n c_{n,k}u^{-k-n}, \tag{10}\label{10}$$ where $$c_{n,k}:=(-1)^n\frac{n!}{k!}\binom{n-1}{k-1}$$ (with $$\binom{n-1}{k-1}=0$$ for $$k=0$$).

So, $$f^{(N)}(x)=f(x)\sum_{n=0}^N\binom Nn a^{N-n} b^n \sum_{k=0}^n (-1)^n\frac{n!}{k!}\binom{n-1}{k-1}(bx)^{-k-n}.$$

Note that, in view of the partial fraction decomposition, \eqref{10} allows one to get comparatively simple expressions for the $$n$$th derivative of any expression of the form $$e^{P(x)/Q(x)}$$, where $$P(x)$$ is any complex polynomial and $$Q(x)$$ is any complex polynomial without multiple roots. Moreover, the no-multiple-roots restriction can be removed, because any complex polynomial with multiple roots and its derivatives can be approximated however closely by a complex polynomial without multiple roots and its derivatives.

Also, writing (say) $$\sin\frac1x=\frac1{2i}(e^{-1/(ix)}-e^{1/(ix)})$$, we can use \eqref{10} to get a rather simple expression for the $$n$$th derivative of $$\sin\frac1x$$.

• I was starting the same computation! This seems the simplest way to connect the $P'_{\lambda,\mu}$'s with something known Commented Apr 23 at 3:26
• @Iosif Pinelis Nice derivation! Note that for $n>0$, $\sum_{k=0}^n \frac{1}{k!}\binom{n-1}{k-1}(bx)^{-k-n}$ $=$ $\frac{1}{n\,(b x)^{n+1}}L_{n-1}^{(1)}(-\frac{1}{bx})$, with the generalized Laguerre polynomials $L$ from my answer. Commented Apr 23 at 6:52
• @PietroMajer : Thank you for your comment. Commented Apr 24 at 1:26
• @FredHucht : Thank you for your comment. Commented Apr 24 at 1:26
• @IosifPinelis: very nice derivation! Thank you Commented Apr 24 at 7:13

It is easy to see that we can write $$\begin{equation*} h(x)=\ln f(x)=-\frac{\lambda(x-\mu)^2}{2\mu^2x}=\frac{\lambda}{2\mu^2}\biggl(2\mu-x-\frac{\mu^2}{x}\biggr). \end{equation*}$$ Hence, its derivatives are $$\begin{equation*} h'(x)=-\frac{\lambda}{2\mu^2}\biggl(1-\frac{\mu^2}{x^2}\biggr) =\frac{\lambda}{2}\biggl(\frac{1}{x^2}-\frac{1}{\mu^2}\biggr) \end{equation*}$$ and $$\begin{equation*} h^{(k+1)}(x)=(-1)^{k}\frac{\lambda}{2}\frac{(k+1)!}{x^{k+2}},\quad k\in\mathbb{N}. \end{equation*}$$ Making use of the identity $$\begin{equation*} B_{n,k}(x_1,x_2,\dotsc,x_{n-k+1}) =\frac{n!}{(n-k)!}\sum_{\ell=0}^{k}\frac{x_1^\ell}{\ell!} B_{n-k,k-\ell}\biggl(\frac{x_2}{2}, \frac{x_3}{3},\dotsc,\frac{x_{n+\ell-2k+2}}{n+\ell-2k+2}\biggr) \end{equation*}$$ for $$n\ge k\ge0$$, where $$B_{n,k}$$ stands for the partial Bell polynomials (also known as the Bell polynomials of the second kind), we obtain \begin{align*} &\quad B_{n,k}\bigl(h'(x),h''(x),\dotsc,h^{(n-k+1)}(x)\bigr)\\ &=\frac{n!}{(n-k)!}\sum_{\ell=0}^{k}\frac{[h'(x)]^\ell}{\ell!} B_{n-k,k-\ell}\biggl(\frac{h''(x)}{2}, \frac{h'''(x)}{3}, \dotsc, \frac{h^{(n+\ell-2k+2)}(x)}{n+\ell-2k+2}\biggr)\\ &=\frac{n!}{(n-k)!}\sum_{\ell=0}^{k}\frac{1}{\ell!} \biggl[\frac{\lambda}{2}\biggl(\frac{1}{x^2}-\frac{1}{\mu^2}\biggr)\biggr]^\ell B_{n-k,k-\ell}\biggl(-\frac{1}{2}\frac{\lambda}{2}\frac{2!}{x^{3}}, \frac{1}{3} \frac{\lambda}{2}\frac{3!}{x^{4}}, \dotsc, \frac{(-1)^{n+\ell-2k+1}}{n+\ell-2k+2}\frac{\lambda}{2}\frac{(n+\ell-2k+2)!}{x^{n+\ell-2k+3}}\biggr)\\ &=\frac{n!}{(n-k)!}\sum_{\ell=0}^{k}\frac{1}{\ell!} \biggl[\frac{\lambda}{2}\biggl(\frac{1}{x^2}-\frac{1}{\mu^2}\biggr)\biggr]^\ell \biggl(\frac{\lambda}{2}\biggr)^{k-\ell} \frac{(-1)^{n-k}}{x^{n+k-2\ell}} B_{n-k,k-\ell}(1!, 2!, \dotsc,(n+\ell-2k+1)!)\\ &=\frac{(-1)^{n-k}}{(n-k)!}\frac{n!}{x^{n+k}}\biggl(\frac{\lambda}{2}\biggr)^{k} \sum_{\ell=0}^{k}\frac{1}{\ell!}\biggl(1-\frac{x^2}{\mu^2}\biggr)^\ell\binom{n-k-1}{k-\ell-1}\frac{(n-k)!}{(k-\ell)!}\\ &=\frac{(-1)^{n+k}}{x^{n+k}}\frac{n!}{k!}\biggl(\frac{\lambda}{2}\biggr)^{k} \sum_{\ell=0}^{k}\binom{k}{\ell}\binom{n-k-1}{k-\ell-1}\biggl(1-\frac{x^2}{\mu^2}\biggr)^\ell \end{align*} for $$n\ge k\ge0$$, where we used the identity $$B_{n,k}\bigl(abx_1,ab^2x_2,\dotsc,ab^{n-k+1}x_{n-k+1}\bigr) =a^kb^n B_{n,k}(x_1,x_2,\dotsc,x_{n-k+1})$$ and the formula $$B_{n,k}(1!,2!,\dotsc,(n-k+1)!)=\binom{n-1}{k-1}\frac{n!}{k!}$$ for $$n\ge k\ge0$$. Further, by the Faa di Bruno formula $$\begin{equation*} \frac{\operatorname{d}^n[f\circ h(x)]}{\operatorname{d} x^n}=\sum_{k=0}^nf^{(k)}(h(x)) B_{n,k}\bigl(h'(x),h''(x),\dotsc,h^{(n-k+1)}(x)\bigr), \quad n\ge0, \end{equation*}$$ we acquire \begin{align*} f^{(n)}(x)&=\sum_{k=0}^{n}\operatorname{e}^{h(x)} B_{n,k}\bigl(h'(x),h''(x),\dotsc,h^{(n-k+1)}(x)\bigr)\\ &=\operatorname{e}^{h(x)}\sum_{k=0}^{n}\frac{(-1)^{n+k}}{x^{n+k}}\frac{n!}{k!}\biggl(\frac{\lambda}{2}\biggr)^{k} \sum_{\ell=0}^{k}\binom{k}{\ell}\binom{n-k-1}{k-\ell-1}\biggl(1-\frac{x^2}{\mu^2}\biggr)^\ell\\ &=f(x)\frac{n!}{x^{2n}}\sum_{k=0}^{n}\frac{(-1)^{n-k}}{k!}\biggl(\frac{\lambda}{2}\biggr)^{k} x^{n-k} \sum_{\ell=0}^{k}\binom{k}{\ell}\binom{n-k-1}{k-\ell-1}\biggl(1-\frac{x^2}{\mu^2}\biggr)^\ell\\ &=\frac{f(x)}{2^n \mu^{2n} x^{2n}}\sum_{k=0}^{n}\lambda^k (-2x)^{n-k} \sum_{\ell=0}^{k}\binom{n}{\ell} \binom{n-k-1}{k-\ell-1} \frac{(n-\ell)!}{(k-\ell)!} \mu^{2(n-\ell)} \bigl(\mu^2-x^2\bigr)^\ell \end{align*} for $$n\ge0$$. In conclusion, we obtain the explicit derivative formula $$\boxed{f^{(n)}(x)=\frac{f(x)}{2^n \mu^{2n} x^{2n}}\sum_{k=0}^{n}\lambda^k (-2x)^{n-k} \sum_{\ell=0}^{k}\binom{n}{\ell} \binom{n-k-1}{k-\ell-1} \frac{(n-\ell)!}{(k-\ell)!} \mu^{2(n-\ell)} \bigl(\mu^2-x^2\bigr)^\ell}$$ for $$n\ge0$$.

References

1. C. A. Charalambides, Enumerative Combinatorics, CRC Press Series on Discrete Mathematics and its Applications. Chapman & Hall/CRC, Boca Raton, FL, 2002.
2. L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Revised and Enlarged Edition, D. Reidel Publishing Co., 1974; available online at https://doi.org/10.1007/978-94-010-2196-8.
3. https://qifeng618.wordpress.com/2018/03/19/some-papers-on-explicit-formulas-of-special-values-for-the-bell-polynomials-of-the-second-kind/.
• Your formula $$f^{(n)}(x)=\frac{e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}}}{2^n \mu^{2 n} x^{2 n}} \sum\limits_{k=1}^n \lambda^k (-2 x)^{n-k} \sum\limits_{\ell=0}^k \binom{n}{\ell} \binom{n-k-1}{k-\ell-1} \frac{(n-\ell)!}{(k-\ell)!} \mu^{2 (n-\ell)} \left(x^2-\mu^2\right)^{\ell}$$ doesnt seem to evaluate correctly for $n>0$ even after I modified the outer sum to start at $k=0$ instead of $k=1$. Commented Apr 22 at 0:09
• @StevenClark I have checked up several times, but didn’t find where the errors locate. I also computed by another approach and verified by Mathematica, I haven’t located where errors appear yet. I hope to correct it tonight. Just wait for a night. Thanks Commented Apr 22 at 12:42
• @StevenClark Numerical computation shows that the derivative formula $$f^{(n)}(x)=\frac{f(x)}{2^n \mu^{2n} x^{2n}}\sum_{k=0}^{n}\lambda^k (-2x)^{n-k} \sum_{\ell=0}^{k}\binom{n}{\ell} \binom{n-k-1}{k-\ell-1} \frac{(n-\ell)!}{(k-\ell)!} \mu^{2(n-\ell)} \bigl(\mu^2-x^2\bigr)^\ell$$ for $n\ge0$ is really not correct. But I cannot find any error in the above proof. What's wrong? What stupid mistake did I make in the above proof? Commented Apr 23 at 2:56
• In fact, the boxed formula in this answer is correct. Commented Apr 24 at 12:20
• No, it is not correct. E.g., $f^{(1)}(x) = (\frac{\lambda}{2x^2}-\frac{\lambda}{2\mu^2})f(x)$, but your formula gives $f^{(1)}(x) = (\frac{\lambda}{x^2}-\frac{\lambda}{2\mu^2})f(x).$ Commented Apr 24 at 12:53

We can give a fairly simple generating function for the polynomials $$P_{\lambda,\mu}$$ from which explicit formulas could be derived.

By Taylor's theorem, $$\sum_{n=0}^\infty f^{(n)}(x) \frac{y^n}{n!} = f(x+y).$$ Writing $$P_{\lambda,\mu,n}$$ for the proposer's $$P_{\lambda,\mu}$$, we then have \begin{align*} \sum_{n=0}P_{\lambda,\mu,n}(x) \frac{y^n}{n!} &= f(x+2\mu^2 x^2 y)/f(x)\\ &= \exp\left(\frac{-\lambda y (x^2-\mu^2+2\mu^2x^3y)}{1+2\mu^2xy}\right). \end{align*} So if we define polynomials $$Q_n(a,b,c)$$ by $$\sum_{n=0}^\infty Q_n(a,b,c)\frac{y^n}{n!} = \exp\left(\frac{ay+by^2}{1+cy}\right)$$ then $$P_{\lambda,\mu,n}(x) = Q_n\bigl(\lambda(\mu^2-x^2), -2\lambda\mu^2 x^3, 2\mu^2 x\bigr).$$

As I mentioned in my Math StackExhange answer Mathematica gives a difference root representation which leads to the formula

$$\frac{\partial^n e^{-\frac{\lambda\, (x-\mu)^2}{2\, \mu^2 x}}}{\partial x^n}=n!\, y(x,n)\tag{1}$$

where $$y(x,n)$$ is defined recursively as

$$y(x,n)=\left\{\begin{array}{cc} e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}} & n=0 \\ e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}} \frac{\lambda \left( \mu^2-x^2\right)}{2 \mu^2 x^2} & n=1 \\ e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}} \frac{\lambda \left(\lambda \mu^4+\lambda x^4-2 \lambda \mu^2 x^2-4 \mu^4 x\right)}{8 \mu^4 x^4} & n=2 \\ -\frac{\left(-\lambda \mu^2+4 \mu^2 (n-3) x+\lambda x^2+8 \mu^2 x\right)\, y(x,n-1)+2 \left( \mu^2+\mu^2 (n-3)+\lambda x\right)\, y(x,n-2)+\lambda\, y(x,n-3)}{2 \mu^2 n x^2} & n\ge 3 \\ \end{array}\right.\tag{2}$$

WolframAlpha gives the more general result

$$\frac{\partial^n e^{f(x)}}{\partial x^n}=e^{f(x)} \sum\limits_{k=0}^n \frac{1}{k!} \sum\limits_{j=0}^k (-1)^j \binom{k}{j} f(x)^j \frac{\partial ^nf(x)^{k-j}}{\partial x^n}\tag{3}$$

and for

$$f(x)=-\frac{\lambda\, (x-\mu)^2}{2\, \mu^2 x}\tag{4}$$

Mathematica simplifies formula (3) above to

$$\frac{\partial^n e^{-\frac{\lambda\, (x-\mu)^2}{2\, \mu^2 x}}}{\partial x^n}=e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}} n! \sum\limits_{k=0}^n \frac{2^{-k}}{k!} \sum\limits_{j=0}^k (-1)^j \binom{k}{j} \left(-\frac{\lambda (x-\mu)^2}{\mu^2 x}\right)^j y_1(x,n)\tag{5}$$

where $$y_1(x,n)$$ is defined recursively as

$$y_1(x,n)=\left\{\begin{array}{cc} \left(-\frac{\lambda (x-\mu )^2}{\mu ^2 x}\right)^{k-j} & n=0 \\ \left(-\frac{\lambda (x-\mu )^2}{\mu ^2 x}\right)^{k-j-1} \frac{(\lambda (j-k) (x-\mu ) (\mu +x))}{\mu ^2 x^2} & n=1 \\ -\frac{(\mu (j-k-n+1)+x (j-k+2 (n-1)))\, y_1(x,n-1)+(j-k+n-2)\, y_1(x,n-2)}{n x (x-\mu )} & n\ge 2 \\ \end{array}\right.\tag{6}$$

which is a slightly simpler recursion than formula (2) above.

An alternative answer is $$\begin{equation*} \frac{\operatorname{d}^n}{\operatorname{d}x^n}\exp\biggl[-\frac{\lambda(x-\mu)^2}{2\mu^2x}\biggr]=\frac{f(x)}{2^n\mu^{2n}x^{2n}}\sum_{k=0}^{n}\binom{n}{k}(-\lambda)^{n-k} 2^{k}\mu^{2k}x^{2(n-k)} \sum_{\ell=0}^k \biggl(-\frac{\lambda}{2}\biggr)^\ell \Biggl[\sum _{j=\ell}^k s(k,j)(-1)^jS(j,\ell)\Biggr] x^{k-\ell} \end{equation*}$$ for $$n\ge0$$, where $$s(n,j)$$ and $$S(j,k)$$ denote the Stirling numbers of the first and second kinds respectively.

References

1. S. Jin, B.-N. Guo, and F. Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, CMES Comput. Model. Eng. Sci. 132 (2022), no. 3, 781--799; available online at https://doi.org/10.32604/cmes.2022.019941.
• Thank you for this derivation and the link with Stirling numbers! Commented Apr 24 at 7:15
• I have verified this answer. Commented Apr 24 at 15:46

In the references below, the derivative formula $$$$\label{exp-frac1x-expans}\tag{QF1} \bigl(\operatorname{e}^{\pm1/t}\bigr)^{(n)} =(-1)^n{\operatorname{e}^{\pm1/t}}\sum_{k=0}^{n}(\pm1)^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0$$$$ was derived alternatively. From \eqref{exp-frac1x-expans}, it follows that $$$$\label{exp-frac1x-alpha}\tag{QF2} \bigl(\operatorname{e}^{\alpha/t}\bigr)^{(n)} =(-1)^n\operatorname{e}^{\alpha/t}\sum_{k=0}^{n}\alpha^{k}\binom{n-1}{k-1}\frac{n!}{k!}\frac1{t^{n+k}}, \quad n\ge0.$$$$ Taking $$\alpha=-\frac{\lambda}{2}$$ in \eqref{exp-frac1x-alpha} gives $$\begin{equation*} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr)\biggr]^{(k)} =(-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{t}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{t^{k+\ell}}, \quad k\ge0. \end{equation*}$$ Therefore, in light of the Leibnitz rule for differentiation, we obtain \begin{gather*} \begin{aligned} f^{(n)}(x)&=\biggl[\exp\biggl(\frac{\lambda}{\mu}\biggr) \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(n)}\\ &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k}\biggl[\exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr)\biggr]^{(n-k)} \biggl[\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr)\biggr]^{(k)} \end{aligned}\\ \begin{aligned} &=\exp\biggl(\frac{\lambda}{\mu}\biggr) \sum_{k=0}^{n}\binom{n}{k} \exp\biggl(-\frac{\lambda}{2\mu^2}x\biggr) \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k\exp\biggl(-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=\exp\biggl(\frac{\lambda}{\mu}-\frac{\lambda}{2\mu^2}x-\frac{\lambda}{2}\frac{1}{x}\biggr) \sum_{k=0}^{n}\binom{n}{k} \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}\\ &=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0. \end{aligned} \end{gather*} In conclusion, we obtain the derivative formula $$\begin{equation*} \boxed{f^{(n)}(x)=f(x)\frac{(-1)^nn!}{2^n\mu^{2n}x^{2n}} \sum_{k=0}^{n}\frac{\lambda^{n-k}}{(n-k)!} 2^{k}\mu^{2k} \sum_{\ell=0}^{k}\frac{(-\lambda)^{\ell}}{(2\ell)!!} \binom{k-1}{\ell-1}x^{2n-k-\ell}, \quad n\ge0.} \end{equation*}$$

References

1. S. Daboul, J. Mangaldan, M. Z. Spivey, and P. J. Taylor, The Lah numbers and the $$n$$th derivative of $$e^{1/x}$$, Math. Mag. 86 (2013), no. 1, 39--47; Available online at http://dx.doi.org/10.4169/math.mag.86.1.039.
2. F. Qi, An explicit formula for the Bell numbers in terms of the Lah and Stirling numbers, Mediterr. J. Math. 13 (2016), no. 5, 2795--2800; available online at https://doi.org/10.1007/s00009-015-0655-7.
3. X.-J. Zhang, F. Qi, and W.-H. Li, Properties of three functions relating to the exponential function and the existence of partitions of unity, Int. J. Open Probl. Comput. Sci. Math. 5 (2012), no. 3, 122--127; available online at https://doi.org/10.12816/0006128.
4. https://qifeng618.wordpress.com/2018/05/10/some-papers-related-to-the-function-e1-x-and-the-lah-numbers/
• Your derivation down to and including your second to last formula $$\frac{\partial^n e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}}}{\partial x^n}=e^{-\frac{\lambda (x-\mu)^2}{2 \mu^2 x}} \sum_{k=0}^{n}\binom{n}{k} \biggl(-\frac{\lambda}{2\mu^2}\biggr)^{n-k} (-1)^k \sum_{\ell=0}^{k}\biggl(-\frac{\lambda}{2}\biggr)^{\ell} \binom{k-1}{\ell-1}\frac{k!}{\ell!}\frac1{x^{k+\ell}}$$ seems to validate, so there must be a mistake in the final step where you moved some of the terms out of the two sums. Commented Apr 24 at 0:41
• @StevenClark I don't find any error in my proof. I numerically verified the boxed derivative formula by the software Mathematica 12 and didn't find it wrong again. I don't know what happened before. Consequently, the boxed derivative formula is correct. By the way, I want to know why you voted down this answer. Commented Apr 24 at 9:06
• I didn't down-vote your answer, I actually like it because it eliminates the recursion in my answer. I didn't actually try your final formula before since you originally indicated it wasn't correct, but I see now it evaluates correctly. I upvoted to cancel the downvote. Commented Apr 24 at 13:26
• @StevenClark I am sorry that I misunderstood your words “ Your derivation down to and including your second to last formula”. Thank you very much for your upvoting this answer. I don’t know why I mistakenly verified the boxed formula several times, perhaps my software Mathematica has bugs or some computer virus intruded into my computer. This time is a lesson for me. Anyway, to the end, we now confirm the boxed formula correct Commented Apr 24 at 23:34