$n$th derivative of $e^{ax^m}$

Question

I am seeking an explicit expression for the $n$th derivative of $e^{ax^m}$.

To achieve this I consider the $n$th coefficient of the Taylor series with $t\rightarrow 0$. \begin{align} e^{a(x+t)^m} &= \sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a(x+t)^m}\frac{t^n}{n!}\\ &=\sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a\left[\sum\limits_{k=0}^{m}\binom{m}{k}x^kt^{m-k}\right]}\frac{t^n}{n!}\\ &=e^{ax^m}\sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a\sum\limits_{k=0}^{m-1}\binom{m-1}{k}x^kt^{m-k-1}}\frac{t^n}{n!} \end{align} But we can also consider the left hand side as \begin{align} e^{a(x+t)^m} &= \sum_{k=0}^\infty\frac{(a(x+t)^m)^k}{k!}\\ &= \sum_{k=0}^\infty\frac{a^k}{k!}\sum_{j=0}^{mk}\binom{mk}{j}x^{j}t^{mk-j}\\ &= \sum_{k=0}^\infty\frac{a^k}{k!}x^{mk}\sum_{j=0}^{mk-1}\binom{mk-1}{j}x^{j}t^{mk-1-j}\\ &= \sum_{k=0}^\infty\frac{(ax^m)^k}{k!}\sum_{j=0}^{mk-1}\frac{(mk-1)!}{j!(mk-1-j)!}x^{j}t^{mk-1-j} \end{align} Although I can't really take it further. My next step would be to extract the $n$th coefficient and, upon comparison to the first formula, obtain an expression for the $n$th derivative. When $m=2$ we obtain the Hermite polynomials. Related to

$n$th derivative of $e^{1/x}$

Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$

and links therein. With thanks to BillyJoe from the comments, Wolfram gives \begin{equation} \frac{d^n}{dx^n}e^{ax^m}=e^{ax^m}x^{-n}\sum^n_{k=0}\sum^k_{j=0}\frac{(-1)^j(ax^m)^k(1-jm+km-n)_n}{j!(k-j)!} \end{equation}

I have also seen that this is equivalent to n exponent rial Riordan array, $B_m=[1,(1+x)^m+1]$. \begin{equation} \frac{d^n}{dx^n}e^{ax^m}=\left( \sum_{k\in \mathbb Z}b_{n,k;m}(ax)^{km-n}\right)e^{ax^m} \end{equation}

Answer 1

By the Faa di Bruno formula \begin{equation}\label{Bruno-Bell-Polynomial} \frac{\operatorname{d}^n}{\operatorname{d}t^n}f\circ h(t) =\sum_{k=0}^nf^{(k)}(h(t)) B_{n,k}\bigl(h'(t),h''(t),\dotsc,h^{(n-k+1)}(t)\bigr), \quad n\ge0, \end{equation} we have \begin{align*} \frac{\operatorname{d}^n}{\operatorname{d}x^n}\operatorname{e}^{ax^m}&= \sum_{k=0}^n \operatorname{e}^{ax^m} B_{n,k}(a\langle m\rangle_1x^{m-1},a\langle m\rangle_2 x^{m-2},\dotsc, a\langle m\rangle_{n-k+1}x^{m-(n-k+1)})\\ &=\operatorname{e}^{ax^m}\sum_{k=0}^n a^k x^{mk-n} B_{n,k}(\langle m\rangle_1, \langle m\rangle_2,\dotsc,\langle m\rangle_{n-k+1})\\ &=\frac{\operatorname{e}^{ax^m}}{x^n}\sum_{k=0}^n (-1)^k\frac{a^k x^{mk}}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\langle m\ell\rangle_n, \end{align*} where we used the formulas \begin{gather} \langle z\rangle_n= \prod_{k=0}^{n-1}(z-k)= \begin{cases} z(z-1)\dotsm(z-n+1), & n\in\mathbb{N};\\ 1,& n=0, \end{cases}\\ B_{n,k}\bigl(\alpha\beta x_1,\alpha \beta^2x_2,\dotsc,\alpha \beta^{n-k+1}x_{n-k+1}\bigr) =\alpha^k\beta^n B_{n,k}(x_1,x_2,\dotsc,x_{n-k+1}),\\ B_{n,k}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc,\langle\alpha\rangle_{n-k+1}) =\frac{(-1)^k}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\langle\alpha\ell\rangle_n. \end{gather}

References

1. Feng Qi, Taylor's series expansions for real powers of two functions containing squares of inverse cosine function, closed-form formula for specific partial Bell polynomials, and series representations for real powers of Pi, Demonstratio Mathematica 55 (2022), no. 1, 710--736; available online at https://doi.org/10.1515/dema-2022-0157.
2. Wei-Shih Du, Dongkyu Lim, and Feng Qi, Several recursive and closed-form formulas for some specific values of partial Bell polynomials, Advances in the Theory of Nonlinear Analysis and its Applications 6 (2022), no. 4, 528--537; available online at https://doi.org/10.31197/atnaa.1170948.
3. Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Applicable Analysis and Discrete Mathematics 16 (2022), no. 2, 427--466; available online at https://doi.org/10.2298/AADM210401017G.
4. Siqintuya Jin, Bai-Ni Guo, and Feng Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, Computer Modeling in Engineering & Sciences 132 (2022), no. 3, 781--799; available online at https://doi.org/10.32604/cmes.2022.019941.

Answer 2

Hint:

Experimentally, the expression is of the form

$$(e^{x^m})^{(n)}=P_{mn}(x^m)x^{(-n)\bmod m}e^{x^m}$$ where $P_{mn}$ is a polynomial of degree $n-\left\lfloor\dfrac nm\right\rfloor$. You can arrange the coefficients in a triangular table, with a recurrence relation between the rows.

$$P_{m.n+1}x^{(-n-1)\bmod m}=\\mx^{m-1}(P'_{mn}(x^m)+P_{mn}(x^m))x^{(-n)\bmod m}+((-n)\bmod m)P_{mn}(x^m)x^{(-n)\bmod m-1}$$

which simplifies differently when $m|n$ or not.