# $n_{th}$ 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 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)

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.

• Thanks for your response. To simplify, the solution that I have shared, is it correct as yours one is a bit complex on the face of it... But I appreciate the reply. – Kashan Jan 24 '18 at 0:24
• @Sjaffry: You're welcome. After work I'll add some comments and an example which might help to clarify things. – Markus Scheuer Jan 24 '18 at 7:52
• @Sjaffry: I've added examples which show your calculations are correct. Good work. :-) Hint: It's preferable to stick at the same notation. In your case the function $L_d$ and the variable $s$. Otherwise you should explicitly mention the notational change. – Markus Scheuer Jan 24 '18 at 15:00
• Thanks for the detailed answer. The only reason I accepted my answer as the solution is that it simplifies the approach, though your approach gives the exact formula for the same thing. I appreciate your time and learned from your response. – Kashan Jan 25 '18 at 21:28
• @Sjaffry: You're welcome! – Markus Scheuer Jan 25 '18 at 21:38

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