# Đạo hàm hàm số đặc biệt

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