Đạo hàm hàm số đặc biệt

Đạo hàm hàm số đặc biệt	Đạo hàm hàm số
$\mathrm{\Gamma }(x)$	$\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^{x-1}{e}^{-t}\ln tdt$
$\mathrm{\Gamma }(x)$	$\mathrm{\Gamma }(x)(\sum _{n=1}^{\mathrm{\infty }}(\ln (1+\frac{1}{n})-\frac{1}{x+n})-\frac{1}{x})=\mathrm{\Gamma }(x)\psi (x)$
$\zeta (x)$	$-\sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n^x}=-\frac{\ln 2}{2^x}-\frac{\ln 3}{3^x}-\frac{\ln 4}{4^x}-\cdots$
$\zeta (x)$	$-\sum _{p\text{ prime}}\frac{p^{-x}\ln p}{(1-p^{-x}{)}^2}\prod _{q\text{ prime},q\ne p}\frac{1}{1-q^{-x}}$
Hàm số	Đạo hàm bậc N
$y=F(G(x))$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=n!\sum _{\{k_m\}}^{}\frac{{\mathrm{d}}^r}{\mathrm{d}{z}^r}F(z){|}_{z=G(x)}\prod _{m=1}^n\frac{1}{k_m!}{(\frac{1}{m!}\frac{{\mathrm{d}}^m}{\mathrm{d}{x}^m}G(x))}^{k_m}$ where $r=\sum _{m=1}^n{k}_m$ and the set $\{k_m\}$ consists of all non-negative integer solutions of the Diophantine equation $\sum _{m=1}^{n}m{k}_m=n$ See: Faà di Bruno's formula, Expansions for nearly Gaussian distributions by S. Blinnikov and R. Moessner
$y=F(x)G(x)$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})\frac{{\mathrm{d}}^{n-k}}{\mathrm{d}{x}^{n-k}}F(x)\frac{{\mathrm{d}}^k}{\mathrm{d}{x}^k}G(x)$ See: General Leibniz rule
$y=x^N$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=\prod _{r=1}^n(N-r+1)x^{N-n}$
$y=[F(x){]}^r$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=r(\genfrac{}{}{0ex}{}{n-r}{n})\sum _{j=0}^n\frac{(-1{)}^j}{r-j}(\genfrac{}{}{0ex}{}{n}{j})[F(x){]}^{r-j}\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}[F(x){]}^j$
$y=B^{Ax}$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=A^n{B}^{Ax}{(\ln B)}^n$
For the case of $B=\exp (1)=e$ (the exponential function), the above reduces to: $y=e^{Ax}$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=A^n{e}^{Ax}$
$y=\ln [F(x)]$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}={\delta }_n\ln [F(x)]+\sum _{j=1}^n\frac{(-1{)}^{j-1}}{j}(\genfrac{}{}{0ex}{}{n}{j})\frac{1}{[F(x){]}^j}\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}[F(x){]}^j$ where ${\delta }_n=\{\begin{array}{cc}1& n=0\\ 0& n\ne 0\end{array}$ is the Kronecker delta.
$y=\sin (Ax+B)$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=A^n\sin (Ax+B+\frac{n\pi }{2})$ Expanding this by the sine addition formula yields a more clear form to use: $\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=A^n[\cos \left(\frac{n\pi }{2}\right)\sin (Ax+B)+\sin \left(\frac{n\pi }{2}\right)\cos (Ax+B)]$
$y=\cos (Ax+B)$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=A^n\cos (Ax+B+\frac{n\pi }{2})$ Expanding by the cosine addition formula: $\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=A^n[\cos \left(\frac{n\pi }{2}\right)\cos (Ax+B)-\sin \left(\frac{n\pi }{2}\right)\sin (Ax+B)]$
$y=\sinh (Ax+B)$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=(-i{A}^n)\sinh (Ax+B+\frac{in\pi }{2})$
$y=\cosh (Ax+B)$	$\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n}=(\pm i{A}^n)\cosh (Ax+B\mp \frac{in\pi }{2})$