数学分析基础知识整理

基本求导公式

$(C{)}^{\prime }=0$
${\left(x^{\mu }\right)}^{\prime }=\mu x^{\mu -1}$

$(\sin x{)}^{\prime }=\cos x$
$(\cos x{)}^{\prime }=-\sin x$
$(\tan x{)}^{\prime }={\sec }^{2}x$
$(\cot x{)}^{\prime }=-{\csc }^{2}x$
$(\sec x{)}^{\prime }=\sec x\tan x$
$(\csc x{)}^{\prime }=-\csc x\cot x$

$\begin{array}{rl}(\arcsin x{)}^{\prime }& =\frac{1}{\sqrt{1-x^2}}\\ (\arccos x{)}^{\prime }& =\frac{-1}{\sqrt{1-x^2}}\\ (\arctan x{)}^{\prime }& =\frac{1}{1+x^2}\\ (\mathrm{arccot}x{)}^{\prime }& =\frac{-1}{1+x^2}\end{array}$

${\left(a^x\right)}^{\prime }=a^x\ln a$
${\left(e^x\right)}^{\prime }=e^x$
${\left({\log }_{a}x\right)}^{\prime }=\frac{1}{x\ln a}$
$(\ln x{)}^{\prime }=\frac{1}{x}$

$(u\pm v{)}^{\prime }=u^{\prime }\pm v^{\prime }$
$(Cu{)}^{\prime }=C{u}^{\prime }$
$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$
${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$

莱布尼茨求导公式

$(u\cdot v{)}^{\prime \prime }=u^{\prime \prime }\cdot v+2u^{\prime }\cdot v^{\prime }+u\cdot v^{\prime \prime }$
$(u\cdot v{)}^{(n)}={\sum }_{k=0}^n{C}_n^k{u}^{(n-k)}{v}^{(k)}$

高阶导数

${\left(a^x\right)}^{(n)}=(\ln a{)}^n\cdot a^x$
${\left(e^x\right)}^{(n)}=e^x$
$(\sin kx{)}^{(n)}=k^n\sin \left(\frac{n\pi }{2}+kx\right)$
$(\cos kx{)}^{(n)}=k^n\cos \left(\frac{n\pi }{2}+kx\right)$
$(\ln x{)}^{(n)}=(-1{)}^{n-1}\frac{(n-1)!}{x^n}$
${\left(\frac{1}{x}\right)}^{(n)}=(-1{)}^n\frac{n!}{x^{n+1}}$
${\left(\frac{1}{x+a}\right)}^{(n)}=(-1{)}^n\frac{n!}{(x+a{)}^{n+1}}$
${\left(x^m\right)}^{(n)}=m(m-1)\cdots (m-n+1)x^{m-n}$
$(u\pm v{)}^{(n)}=u^{(n)}\pm v^{(n)}$

复合函数求导法则

 设函数$z=f(u,v)$可微$u=u(x,y),v=v(x,y)$具有一阶偏导数，并且它们可以构成 ，$z$关于$(x,y)$在某区域D内的复合函数，则在D内有复合函数求导法则：
$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial x}$

$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\cdot \frac{\partial v}{\partial y}$

隐函数求导法则

  设函数 $F(x,y,z)$ 在点 $P_0\left(x_0,y_0,z_0\right)$ 的某邻域内有连续偏导数, 并且 $F\left(x_0,y_0,z_0\right)=0$, $F_z^{\prime }\left(x_0,y_0,z_0\right)\ne 0,$ 则方程 $F\left(x_0,y_0,z_0\right)=0$ 在点 $P_0$ 的某一邻域内恒能确定唯一的连续函数 $z=f(x,y)$满足:

• $z_0=f\left(x_0,y_0\right)$
• $F(x,y,f(x,y))\equiv 0$
• $z=f(x,y)$ 具有连续偏导数
  则:
  $\frac{\partial z}{\partial x}=-\frac{F_x^{\prime }(x,y,z)}{F_z^{\prime }(x,y,z)}$
  $\frac{\partial z}{\partial y}=-\frac{F_y^{\prime }(x,y,z)}{F_z^{\prime }(x,y,z)}$

基本积分公式

$\int kdx=kx+C$
$\int x^{\mu }dx=\frac{1}{\mu +1}\cdot x^{\mu +1}+C(\mu \ne -1)$
$\int \frac{dx}{x}=\ln |x|+C$

$\int \cos xdx=\sin x+C$
$\int \sin xdx=-\cos x+C$
$\int \frac{dx}{{\cos }^{2}x}=\int {\sec }^{2}xdx=\tan x+C$
$\int \frac{dx}{{\sin }^{2}x}=\int {\csc }^{2}xdx=-\cot x+C$
$\int \sec x\tan xdx=\sec x+C$
$\int \csc x\cot xdx=-\csc x+C$

$\int \frac{dx}{1+x^2}=\arctan x+C$
$\int \frac{dx}{\sqrt{1-x^2}}=\arcsin x+C$
$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\arctan \frac{x}{a}+C$
$\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin \frac{x}{a}+C$
$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\ln \left|\frac{x-a}{x+a}\right|+C$
$\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \left|\frac{x+a}{x-a}\right|+C$
$\int \frac{dx}{\sqrt{x^2+a^2}}=\ln \left(x+\sqrt{x^2+a^2}\right)+C$
$\int \frac{dx}{\sqrt{x^2-a^2}}=\ln \left|x+\sqrt{x^2-a^2}\right|+C$
$\int \sqrt{x^2-x^2}dx=-\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\arcsin \frac{x}{a}+C$

$\int \tan xdx=\int \frac{\sin x}{\cos x}dx=-\int \frac{d\cos x}{\cos x}=-\ln |\cos x|+C$
$\int \cot xdx=\ln |\sin x|+C$
$\int \csc xdx=\ln \left|\tan \frac{x}{2}\right|+C=\ln |\csc x-\cot x|+C$
$\int \sec xdx=\ln |\sec x+\tan x|+C$

$\int e^{x}dx=e^x+C$
$\int a^{x}dx=\frac{1}{\ln a}{a}^x+C$

万能代换

$\begin{array}{rl}u& =\tan \frac{x}{2}\\ \sin x& =\frac{2u}{1+u^2}\\ \cos x& =\frac{1-u^2}{1+u^2}\\ dx& =\frac{2}{1+u^2}du\end{array}$

万能代换可以将三角函数换成有理分式的形式来进行积分。

$\{\begin{array}{l}\sqrt{a^2-x^2}\to x=a\sin t\\ \sqrt{a^2+x^2}\to x=\mathrm{atan}t\\ \sqrt{x^2-a^2}\to x=\mathrm{asec}t\end{array}$

分部积分

$\int udv=uv-\int vdu$

级数定积分转换

$\{\begin{array}{l}{\sum }_{i=1}^{\infty }f\left(\frac{i}{n}\right)\frac{1}{n}={\int }_0^{1}f(t)dt\\ {\sum }_{i=1}^{\infty }f\left(\frac{x\cdot i}{n}\right)\frac{x}{n}={\int }_0^{x}f(t)dt\\ {\sum }_{i=1}^{\infty }f\left(a+\frac{(b-a)\cdot i}{n}\right)\frac{b-a}{n}={\int }_a^{b}f(t)dt\end{array}$

变限积分函数求导

$F(x)={\int }_{{\phi }_1(x)}^{{\phi }_2(x)}f(t)dt$
$F^{\prime }(x)=\frac{d}{dx}\left[{\int }_{{\phi }_1(x)}^{{\phi }_2(x)}f(t)dt\right]=f\left[{\phi }_2(x)\right]{\phi }_2^{\prime }(x)-f\left[{\phi }_1(x)\right]{\phi }_1^{\prime }(x)$

二重积分

二重积分可以化为累次积分的定积分，表示曲顶柱体的体积。

${\iint }_{D}f(x,y)d\sigma$

$={\int }_{a_x}^{b_x}dx{\int }_{a_y}^{b_y}f(x,y)dy$

$={\int }_{a_{\theta }}^{b_{\theta }}d\theta {\int }_{a_r}^{b_r}f(r\cos \theta ,r\sin \theta )rdr$
${\sum }_{i=1}^{\infty }{\sum }_{j=1}^{\infty }f\left(\frac{i}{n},\frac{j}{n}\right)\frac{1}{n^2}={\int }_0^1{\int }_0^{1}f(x,y)dxdy$

三重积分

三重积分可以化为累次积分的定积分，表示空间区域 以密度$\rho =f(x,y,z)$的质量。
$\begin{array}{rl}& {\iiint }_{\Omega }f(x,y,z)dv\\ =& {\int }_{a_x}^{b_x}dx{\int }_{a_y}^{b_y}dy{\int }_{a_z}^{b_z}f(x,y,z)dz\\ =& {\int }_{a_z}^{b_z}dz{\iint }_{D}f(x,y,z)d\sigma \\ =& {\iiint }_{\Omega }f(r\cos \theta ,r\sin \theta ,z)rdrd\theta \\ =& {\iiint }_{\Omega }f(r\sin \phi \cos \theta ,r\sin \phi \sin \theta ,r\cos \phi )\cdot r^2\sin \phi drd\phi d\theta \end{array}$

常用泰勒公式

$f(x)=f\left(x_0\right)+f^{\prime }\left(x_0\right)\left(x-x_0\right)+\frac{f^{\prime \prime }\left(x_0\right)}{2!}{\left(x-x_0\right)}^2+\cdots +\frac{f^{(n)}\left(x_0\right)}{n!}{\left(x-x_0\right)}^n+R_n(x)$
$R_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}{\left(x-x_0\right)}^{(n+1)}\xi \in \left(x,x_0\right)$
$R_n(x)=o\left[{\left(x-x_0\right)}^n\right]$
$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots +(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}+o\left(x^{2n+1}\right)$
$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +(-1{)}^n\frac{x^{2n}}{(2n)!}+o\left(x^{2n}\right)$
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!}+o\left(x^n\right)$
$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots +(-1{)}^{n-1}\frac{x^n}{n}+o\left(x^n\right)$
$\tan x=x+\frac{x^3}{3}+o\left(x^3\right)$
$\arcsin x=x+\frac{x^3}{3!}+o\left(x^3\right)$
$\arctan x=x-\frac{x^3}{3}+o\left(x^3\right)$
$(1+x{)}^a=1+ax+\frac{a(a-1)}{2!}{x}^2+o\left(x^2\right)$

重要幂级数展开式

$e^x={\sum }_{n=0}^{\infty }\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!}+\cdots ,-\infty <x<+\infty$
$\frac{1}{1+x}={\sum }_{n=0}^{\infty }(-1{)}^n{x}^n=1-x+x^2-\cdots +(-1{)}^n{x}^n+\cdots ,-1<x<1$
$\frac{1}{1-x}={\sum }_{n=0}^{\infty }{x}^n=1+x+x^2+\cdots +x^n+\cdots ,-1<x<1$
$\ln (1+x)={\sum }_{n=1}^{\infty }(-1{)}^{n-1}\frac{x^n}{n}=x-\frac{x^2}{2}+\cdots +(-1{)}^{n-1}\frac{x^n}{n}+\cdots ,-1<x⩽1$
$\sin x={\sum }_{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\cdots +(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}+\cdots ,-\infty <x<+\infty$
$\cos x={\sum }_{n=0}^{\infty }(-1{)}^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\cdots +(-1{)}^n\frac{x^{2n}}{(2n)!}+\cdots ,-\infty <x<+\infty$

无穷级数的和函数

$\{\begin{array}{ll}{\sum }_{n=1}^{\infty }n{x}^{n-1}=\frac{1}{(1-x{)}^2}& (-1<x<1)\\ {\sum }_{n=1}^{\infty }\frac{1}{n}{x}^n=-\ln (1-x)& (-1⩽x<1)\end{array}$

傅里叶级数

$\begin{array}{rl}f(x)& =\frac{a_0}{2}+\sum _{n=1}^{\infty }\left(a_n\cos \frac{n\pi x}{l}+b_n\sin \frac{n\pi x}{l}\right)\\ a_0& =\frac{1}{l}{\int }_{-l}^{l}f(x)dx\\ a_n& =\frac{1}{l}{\int }_{-l}^{l}f(x)\cos \frac{n\pi x}{l}dx\\ b_n& =\frac{1}{l}{\int }_{-l}^{l}f(x)\sin \frac{n\pi x}{l}dx\end{array}$