$n_{th}$ derivative of function? I have to take the nth derivative of a function (below). 
The research paper, which I have taken it from calls it the composite function. Any idea how to find the solution?
$$L_d(s) = exp\{-C_d s^{2/\alpha_N}\}$$
I can take as many derivates as I want manually, but cannot find the $nth$ derivate or any formula for it, For example see below:
$$F' = -\dfrac{2C_dx^{\frac{2}{\alpha_N}-1}\mathrm{e}^{-C_dx^\frac{2}{\alpha_N}}}{\alpha_N}$$
$$F'' = \dfrac{2C_dx^{\frac{2}{\alpha_N}-2}\left(2C_dx^\frac{2}{\alpha_N}+\alpha_N-2\right)\mathrm{e}^{-ax^\frac{2}{\alpha_N}}}{\alpha_N^2}$$
Any guidance would be appreciated for $F^n$
 A: A formula to calculate the $n$-th derivative of composite functions is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:

Hoppe Form of Generalized Chain Rule
Let $D_s$ represent differentiation with respect to $s$ and $z=z(s)$. Hence $D^n_s g(z)$ is the $n$-th derivative of $g$ with respect to $s$. The following holds true
  \begin{align*}
D_s^n g(z)=\sum_{k=0}^nD_z^kg(z)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}z^{k-j}D_s^nz^j
\end{align*}
  In the special case
  \begin{align*}
g(z(s))=e^{z(s)}
\end{align*}
  we have $$D_z^kg(z)=D_z^k e^z=e^z$$ and obtain
  \begin{align*}
D_s^ne^z=e^z\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}z^{k-j}D_s^nz^s\tag{1}
\end{align*}

We consider 
\begin{align*}
z=z(s)=as^b\tag{2}
\end{align*}
which corresponds to $L_d(s) = exp\{-C_d s^{2/\alpha_N}\}$ with $a=-C_d$ and $b=\frac{2}{\alpha_N}$.

From (1) and (2) we obtain for $n>0$:
  \begin{align*}
\color{blue}{D_s^n\exp\left(as^b\right)}
&=\exp\left(as^b\right)\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}\left(as^b\right)^{k-j}D_s^n\left[\left(as^{b}\right)^j\right]\\
&=\exp\left(as^b\right)\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}a^{k-j}s^{b(k-j)}D_s^n\left(a^js^{bj}\right)\\
&=\exp\left(as^b\right)\sum_{k=0}^n\frac{(-1)^k}{k!}a^k\sum_{j=0}^k(-1)^j\binom{k}{j}s^{b(k-j)}D_s^n\left(s^{bj}\right)\\
&\color{blue}{=\exp\left(as^b\right)\sum_{k=1}^n\frac{(-a)^k}{k!}\left(\sum_{j=1}^k(-1)^j\binom{k}{j}{(bj)}^{\underline{n}}\right)s^{bk-n}}\tag{3}\\
\end{align*}
  with $m^{\underline{n}}=m(m-1)\cdots(m-n+1)$ the falling factorial.

In the last line the inner sum contains a factor $j$ if $n>0$. We can therefore start with the lower bound $j=1$ and so also with $k=1$.
Example: $n=1$:
We obtain from (3)
\begin{align*}
  \exp&\left(as^b\right)\sum_{k=1}^1\frac{(-a)^k}{k!}\left(\sum_{j=1}^k(-1)^j\binom{k}{j}bj\right)s^{bk-1}\\
  &=\exp\left(as^b\right)\left(\frac{(-a)^1}{1}\left[(-1)\binom{1}{1}b\right]\right)s^{b-1}\\
  &=\exp\left(as^b\right)abs^{b-1}\tag{4}
  \end{align*}
Substituting $a=-C_d,b=\frac{2}{\alpha_N}$ in (4) gives
\begin{align*}
  \color{blue}{\exp\left(-\frac{2C_d}{\alpha_N}\right)\left(-\frac{2C_d}{\alpha_N}\right)s^{\frac{2}{\alpha_N}-1}}
  \end{align*}
in accordance with OPs calculation.
Example: $n=2$:
We obtain from (3)
\begin{align*}
  \exp&\left(as^b\right)\sum_{k=1}^2\frac{(-a)^k}{k!}\left(\sum_{j=1}^k(-1)^j\binom{k}{j}bj(bj-1)\right)s^{bk-2}\\
  &=\exp\left(as^b\right)\left(\frac{(-a)^1}{1}\left[(-1)\binom{1}{1}b(b-1)\right]\right.s^{b-2}\\
  &\qquad\qquad\qquad\left.+\frac{(-a)^2}{2!}\left[(-1)\binom{2}{1}b(b-1)+\binom{2}{2}2b(2b-1)\right]s^{2b-2}\right)\\
  &=\cdots\\
  &=\exp\left(as^b\right)\left(ab(b-1)s^{b-2}+a^2b^2s^{2b-2}\right)\tag{5}
  \end{align*}
Substituting $a=-C_d,b=\frac{2}{\alpha_N}$ in (5) gives
\begin{align*}
  \exp&\left(-\frac{2C_d}{\alpha_N}\right)\left(\left(-C_d\right)\frac{2}{\alpha_N}\left(\frac{2}{\alpha_N-1}\right)s^{\frac{2}{\alpha_N-2}}+C_d^2\frac{4}{\alpha_N^2}s^{\frac{4}{\alpha_N-2}}\right)\\
  &=\color{blue}{\exp\left(as^b\right)\frac{2C_d}{\alpha_N^2}\left(\alpha_N-2+2C_ds^{\frac{2}{\alpha_N}}\right)}
  \end{align*}
in accordance with OPs calculation. 
A: Can we do the following (is it correct?)
Since we have the formula: 
$f(x) = e^x = \sum^{\infty}_{j=0} \frac{x^j}{j!}$
Hence, 
$$f(x) = e^{Ax^{2/b}} = \sum^{\infty}_{j=0} \frac{(Ax^{2/b})^j}{j!}$$
$$f(x) = \sum^{\infty}_{j=0} \frac{A^j}{j!}(x^{2j/b})$$
Then, let $z = x^{p\ j}$, where $p = 2/b$
$$ Z^1 = \frac{d}{dx}(x^{p \ j}) = (pj)x^{p\ j-1}$$
$$ Z^2 = \frac{d}{dx}(Z^1) = (pj)(p j-1)x^{p\ j-2}$$
$$ Z^3 = (pj)(p j-1)(p j-2)x^{p\ j-3}$$
$$ Z^3 = (pj)(p j-1)(p j-2)(p j-3)x^{p\ j-4}$$
Then
$$ Z^n = (pj)(pj-1)(pj-2)...(pj-n+1)x^{p\ j-n}$$
So, $F^n = \frac{f^n}{f^nx}$ is:
$$ F^n = \sum^{\infty}_{j=0} \frac{A^j}{j!} (pj)(pj-1)(pj-2)...(pj-n+1)x^{p\ j-n}$$
$$ F^n = \sum^{\infty}_{j=0} \frac{A^j}{j!} x^{p\ j-n} (pj)_n$$
Where $(.)_n$ is the falling factorial
