考研高等数学公式总结（一）

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基础预备知识

三角函数

$\csc \alpha =\frac{1}{\sin \alpha }$

$\sec \alpha =\frac{1}{\cos \alpha }$

$\cot \alpha =\frac{1}{\tan \alpha }$

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${\sin }^2\alpha +{\cos }^2\alpha =1$

$1+{\tan }^2\alpha ={\sec }^2\alpha$

$1+{\cot }^2\alpha ={\csc }^2\alpha$

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诱导公式

$\theta =\frac{k\pi }{2}+\alpha$ 奇变偶不变，符号看象限

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倍角公式

$\sin 2\alpha =2\sin \alpha \cos \alpha$

$\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha =1-2{\sin }^2\alpha =2{\cos }^2\alpha -1$

$\sin 3\alpha =-4{\sin }^3\alpha +3\sin \alpha$

$\cos 3\alpha =4{\cos }^3\alpha -3\cos \alpha$

$\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$

$\cot 2\alpha =\frac{{\cot }^2\alpha -1}{2\cot \alpha }$

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半角

${\sin }^2\frac{\alpha }{2}=\frac{1}{2}(1-\cos \alpha )$

${\cos }^2\frac{\alpha }{2}=\frac{1}{2}(1+\cos \alpha )$

$\sin \frac{\alpha }{2}=\pm \sqrt{\frac{1-\cos \alpha }{2}}$

$\cos \frac{\alpha }{2}=\pm \sqrt{\frac{1+\cos \alpha }{2}}$

$\tan \frac{\alpha }{2}=\frac{1-\cos \alpha }{\sin \alpha }=\frac{\sin \alpha }{1+\cos \alpha }=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}$

$\cot \frac{\alpha }{2}=\frac{\sin \alpha }{1-\cos \alpha }=\frac{1+\cos \alpha }{\sin \alpha }=\pm \sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }}$

和差

$\sin (\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$

$\cos (\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$

$\tan (\alpha \pm \beta )=\frac{\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }$

$\cot (\alpha \pm \beta )=\frac{\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }$

积化和差

记住第一个，其他用诱导公式推

$\sin \alpha \cos \beta =\frac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )]$

$\cos \alpha \sin \beta =\frac{1}{2}[\sin (\alpha +\beta )-\sin (\alpha -\beta )]$

$\cos \alpha \cos \beta =\frac{1}{2}[\cos (\alpha +\beta )+\cos (\alpha -\beta )]$

$\sin \alpha \sin \beta =\frac{1}{2}[\cos (\alpha -\beta )-\cos (\alpha +\beta )]$

和差化积

记住第一个，其他用诱导公式推

$\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$

$\sin \alpha -\sin \beta =2\cos \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$

$\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$

$\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$

万能公式

$u=\tan \frac{x}{2}\Rightarrow \sin x=\frac{2u}{1+u^2},\cos x=\frac{1-u^2}{1+u^2}$

因式分解公式

$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$

$(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

$a^3+b^3=(a+b)(a^2-ab+b^2)$

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+a{b}^{n-2}+b^{n-1})$

$n是正偶数，a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+...+a{b}^{n-2}-b^{n-1})$

$n是正奇数，a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+...-a{b}^{n-2}+b^{n-1})$

$(a+b{)}^n={\sum }_{k=0}^n{C}_n^k{a}^{n-k}{b}^k$

常用不等式

$\sqrt{ab}⩽\frac{a+b}{2}⩽\sqrt{\frac{a^2+b^2}{2}}(a,b>0)$

$\sqrt[3]{abc}⩽\frac{a+b+c}{3}⩽\sqrt{\frac{a^2+b^2+c^2}{3}}(a,b,c>0)$

$a<x<b,c<y<d\Rightarrow \frac{a}{d}<\frac{x}{y}<\frac{b}{c}$

$\sin x<x<\tan x(0<x<\frac{\pi }{2})$

$arc\tan x⩽x⩽arc\sin x(0⩽x⩽1)$

$e^x⩾x+1$

$\ln x⩽x-1(x>0)$

$\frac{1}{1+x}<\ln (1+\frac{1}{x})<\frac{1}{x}(x>0)$

$\frac{x}{1+x}<\ln (1+x)<x(x>0)$

放缩法

$$\begin{gathered} \sum _{i=1}^n{u}_i=u_1+u_2+...+u_n \\ 当n无穷大，则n\cdot u_{min}⩽\sum _{i=1}^n{u}_i⩽n\cdot u_{max} \\ 当n为有限数，则1\cdot u_{max}⩽\sum _{i=1}^n{u}_i⩽n\cdot u_{max} \end{gathered}$$

泰勒公式

$$f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+...+\frac{1}{n!}{f}^{(n)}(x_0)(x-x_0{)}^n+\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}，\xi \in (x,x_0)\text{\hspace{1em}(带拉格朗日余项的泰勒公式)}$$

$$f(x)=f(x_0)+f^{\prime }(x_0)(x-x_0)+...+\frac{1}{n!}{f}^{(n)}(x_0)(x-x_0{)}^n+o((x-x_0{)}^n)\text{\hspace{1em}(带佩亚诺余项的泰勒公式)}$$

$$泰勒公式中令x=0的展开，可带拉格朗日或佩亚诺余项，有时将\xi =\theta x,\theta \in (0,1)\text{\hspace{1em}(麦克劳林公式)}$$

$$f(x,y)=f(x_0,y_0)+(f_x^{\prime },f_y^{\prime }{)}_{X_0}\left|\begin{array}{c}\Delta x\\ \Delta y\end{array}\right|+\frac{1}{2!}(\Delta x\Delta y){\left|\begin{array}{cc}{f}_{xx}^{\prime \prime }& f_{xy}^{\prime \prime }\\ f_{yx}^{\prime \prime }& f_{yy}^{\prime \prime }\end{array}\right|}_{X_0}\left|\begin{array}{c}\Delta x\\ \Delta y\end{array}\right|+R\text{\hspace{1em}(多元函数泰勒公式)}$$

$$\begin{array}{rl}& e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3)=\sum _{n=0}^{\infty }\frac{x^n}{n!}\\ & \frac{1}{1+x}=1-x+x^2-x^3+o(x^3)=\sum _{n=0}^{\infty }(-1{)}^n{x}^n,|x|<1\\ & \frac{1}{1-x}=1+x+x^2+x^3+o(x^3)=\sum _{n=0}^{\infty }{x}^n,|x|<1\\ & \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)=\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{x^n}{n},-1<x⩽1\\ & \sin x=x-\frac{x^3}{3!}+o(x^3)=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}\\ & \cos x=1-\frac{x^2}{2}+o(x^3)=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n}}{(2n)!}\\ & (1+x{)}^a=1+ax+\frac{a(a-1)}{2!}{x}^2+...+\frac{a(a-1)...(a-n+1)}{n!}{x}^n+...\\ & \tan x=x+\frac{x^3}{3}+o(x^3)\\ & arc\sin x=x+\frac{x^3}{6}+o(x^3)\\ & arc\tan x=x-\frac{x^3}{3}+o(x^3)=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{2n+1},|x|⩽1\\ & \frac{e^x+e^{-x}}{2}=\sum _{n=0}^{\infty }\frac{x^{2n}}{(2n)!}\\ & \frac{e^x-e^{-x}}{2}=\sum _{n=0}^{\infty }\frac{x^{2n+1}}{(2n+1)!}\\ & -\ln (1-x)=\sum _{n=1}^{\infty }\frac{x^n}{n},1⩽x<1\\ & \frac{x}{(1-x{)}^2}=\sum _{n=1}^{\infty }n{x}^n,-1<x<1\end{array}$$

常用无穷小

sin ⁡   x ∼ tan ⁡   x ∼ a r c sin ⁡   x ∼ a r c tan ⁡   x ∼ x ln ⁡ ( 1 + x ) ∼ e x − 1 ∼ x a x − 1 ∼ x ln ⁡ a 1 − cos ⁡   x ∼ 1 2 x 2 ( 1 + 1 x ) x = e ( x → + ∞ ) ( 1 + x ) α − 1 ∼ α x \begin{aligned} &\\ & \sin \ x \sim \tan \ x \sim arc\sin \ x \sim arc\tan \ x \sim x \\ & \ln (1+x) \sim e^x -1 \sim x \\ & a^x-1 \sim x \ln a \\ & 1-\cos \ x \sim \cfrac 12 x^2 \\ &(1+\cfrac 1x)^x = e(x \rightarrow + \infty)\\ & (1+x)^{\alpha}-1 \sim \alpha x \end{aligned} ​sin x∼tan x∼arcsin x∼arctan x∼xln(1+x)∼ex−1∼xax−1∼xlna1−cos x∼21​x2(1+x1​)x=e(x→+∞)(1+x)α−1∼αx​

基本求导公式

$(x^{\alpha }{)}^{\prime }=a{x}^{\alpha -1}$

$(a^x{)}^{\prime }=a^x\ln a$

$(e^x{)}^{\prime }=e^x$

$({\log }_{a}x{)}^{\prime }=\frac{1}{x\ln a}(a>0,a\ne 1)$

$(\ln x{)}^{\prime }=\frac{1}{x}$

$(\sin x{)}^{\prime }=\cos x$

$(\cos x{)}^{\prime }=-\sin x$

$(arc\sin x{)}^{\prime }=\frac{1}{\sqrt{1-x^2}}$

$(arc\cos x{)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}$

$(\tan x{)}^{\prime }={\sec }^{2}x$

$(\cot x{)}^{\prime }=-{\csc }^{2}x$

$(arc\tan x{)}^{\prime }=\frac{1}{1+x^2}$

$(arc\cot x{)}^{\prime }=-\frac{1}{1+x^2}$

$(\sec x{)}^{\prime }=\sec x\tan x$

$(\csc x{)}^{\prime }=-\csc x\cot x$

$[\ln (x+\sqrt{x^2+1}){]}^{\prime }=\frac{1}{\sqrt{x^2+1}}$

$[\ln (x+\sqrt{x^2-1}){]}^{\prime }=\frac{1}{\sqrt{x^2-1}}$

高阶求导公式

$$\frac{d^{2}y}{d{x}^2}=\frac{d\frac{dy}{dx}}{dx}=\frac{d(\frac{dy}{dx}/dt)}{dx/dt}=\frac{{\psi }^{\prime \prime }(t){\varphi }^{\prime }(t)-{\psi }^{\prime }(t){\psi }^{\prime \prime }(t)}{[{\psi }^{\prime }(t){]}^3}\text{\hspace{1em}(参数方程二阶)}$$

$$y_{xx}^{\prime \prime }=\frac{d(\frac{dy}{dx})}{dx}=\frac{d(\frac{1}{x_y^{\prime }})}{dy}.\frac{1}{x_y^{\prime }}=\frac{-x_{yy}^{\prime \prime }}{(x_y^{\prime }{)}^3}\text{\hspace{1em}(反函数二阶)}$$

$$(uv{)}^{(n)}=\sum _{k=0}^n{C}_n{u}^{(n-k)}{v}^{(k)}\text{\hspace{1em}(莱布尼茨公式)}$$

$(a^x{)}^{(n)}=a^x(\ln a{)}^n$

$(e^x{)}^{(n)}=e^x$

$(\sin kx{)}^{(n)}=k^n\sin (kx+n\ast \frac{\pi }{2})$

$(\cos kx{)}^{(n)}=k^n\cos (kx+n\ast \frac{\pi }{2})$

$(\frac{1}{x+a}{)}^{(n)}=\frac{(-1{)}^{n}n!}{(x+a{)}^{n+1}}$

$(\ln x{)}^{(n)}=(-1{)}^{n-1}\frac{(n-1)!}{x^n}$

$[\ln (1+x){]}^{(n)}=(-1{)}^{n-1}\frac{(n-1)!}{(1+x{)}^n}$

$[(x+x_0{)}^m{]}^{(n)}=m(m-1)(m-2)...(m-n+1)(x+x_0{)}^{m-n}$

中值定理

定理	条件	结论
有界与最值定理	$f(x)在[a,b]上连续$	$m⩽f(x)⩽M$
介值定理	$f(x)在[a,b]上连续$，$m⩽\mu ⩽M$	存在$\xi \in [a,b]$，使得$f(\xi )=\mu$
平均值定理	$f(x)在[a,b]上连续$，$a<x_1<x_2<...<x_n<b$	在$[x_1,x_n]$内存在$\xi$,使$f(\xi )=\frac{f(x_1)+f(x_2)+...+f(x_n)}{n}$
零点定理	$f(x)在[a,b]上连续，$$f(a)\ast f(b)<0$	存在$\xi \in [a,b]$，使得$f(\xi )=0$
费马定理	$f(x)$在$x_0$处：可导+取极值	$f^{\prime }(x_0)=0$
罗尔定理	$f(x)$满足：$[a,b]$连续+$(a,b)$可导+$f(a)=f(b)$	存在$\xi \in (a,b)$，使得$f^{\prime }(\xi )=0$
拉格朗日中值定理	$f(x)$满足：$[a,b]$连续+$(a,b)$可导	存在$\xi \in (a,b)$，使得$f(b)-f(a)=f^{\prime }(\xi )(b-a)$
柯西中值定理	$f(x)$满足：$[a,b]$连续+$(a,b)$可导=$g^{\prime }(x)\ne 0$	存在$\xi \in (a,b)$，使得$\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f^{\prime }(\xi )}{g^{\prime }(\xi )}$
积分中值定理	$f(x)$在$[a,b]$上连续	存在$\xi \in [a,b]$或者$\xi \in (a,b)[加强形式]$，使得${\int }_a^{b}f(x)dx=f(\xi )(b-a)$
二重积分中值定理	$f(x,y)$在$D$上连续	存在$(\xi ,\eta )\in D$,使得${\iint }_{D}f(x,y)d\sigma =S_D\times f(\xi ,\eta )$

常见辅助函数构造

函数形式	简明形式	构造函数（原函数）$F(x)=$
$f^{\prime }(x)+g(x)f(x)=h(x)$	$f^{\prime }+gf=h$	$F(x)=f(x)e^{\int g(x)dx}-\int h(x)e^{\int g(x)dx}dx$
$2f(x)\ast f^{\prime }(x)$	$f{f}^{\prime }$	$f^2(x)$
$[f^{\prime }(x){]}^2+f(x)f^{\prime \prime }(x)$	$[f^{\prime }{]}^2+f{f}^{\prime \prime }$	$f(x)f^{\prime }(x)$
$f^{\prime }(x)+f(x){\varphi }^{\prime }(x)$	$f^{\prime }+\varphi f$​	$f(x)e^{\varphi (x)}$
$f^{\prime }(x)+f(x)$	$f^{\prime }+f$	$f(x)e^x$
$f^{\prime }(x)-f(x)$	$f^{\prime }-f$	$f(x)e^{-x}$
$f^{\prime }(x)+kf(x)$	$f^{\prime }+kf$	$f(x)e^{kx}$
$f(x)$	$f$	${\int }_a^{x}f(t)dt$
$\frac{f^{\prime }(x)}{f(x)}$	$\frac{f^{\prime }}{f}$	$\ln f(x)$
$f^{\prime }(x)x-f(x)$	$f^{\prime }x-f$	$\frac{f(x)}{x}$
$f^{\prime \prime }(x)f(x)-[f^{\prime }(x){]}^2$​	$f^{\prime \prime }f-[f^{\prime }{]}^2$​	$\frac{f^{\prime }(x)}{f(x)}$或$\ln f(x)$的二阶导用于研究凹凸性
$u^{\prime \prime }+2u^{\prime }{v}^{\prime }+v^{\prime \prime }$		$(uv{)}^{\prime \prime }$
${\int }_a^{b}f(x)dx$		${\int }_a^{x}f(t)dt$