记不住的高数公式

这你都记不住？今晚给我复习到两点！

基本常识

不等式

\( |\,\,|\, a \,|\, - \, |\, b \,|\,\,| \leq |\, a \pm b \,| \leq |\,a \,| + |\, b \,| \)

\( \frac{2}{\frac{1}{a}+\frac{1}{b}}\, \leq \, \sqrt{ab} \, \leq \, \frac{a+b}{2}\, \leq \, \sqrt{\frac{a^2+b^2}{2}} \)

\( \sqrt[3]{abc}\, \leq \, \frac{a+b+c}{3}\, \leq \, \frac{a^2+b^2+c^2}{3} \)

\( \sqrt[n]{a_1 a_2 \cdots a_n}\, \leq \, \frac{a_1+a_2+ \cdots + a_n}{n} \)

立方公式

\( a^3+b^3 = (a+b)\,(a^2-ab+b^2) \)\( a^3-b^3 = (a-b)\,(a^2+ab+b^2) \)

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

(a+b)^n = C_n^k \, a^0 b^n + C_n^1 \, a^1 b^{n-1} + C_n^2 \, a^2 b^{n-2} + \cdots + C_n^n \, a^n b^0 = \sum_{k=0}^n \; C_n^k \, a^k b^{n-k}C_n^k = \frac{n!}{k! \, (n-k)!}

三角变换

\cos 2\alpha = \cos ^2 \alpha - \sin ^2 \alpha = 2 \cos ^2 \alpha - 1 = 1 - 2 \sin ^2 \alpha\tan 2\alpha = \frac{2\tan \alpha}{1- \tan ^2 \alpha}\sin ^2 \alpha = \frac{1- \cos 2\alpha}{2}\cos ^2 \alpha = \frac{1+ \cos 2\alpha}{2}\tan ^2 \alpha = \frac{1- \cos 2\alpha}{1+ \cos 2\alpha}

分子有理化

\( \sqrt[]{a} - \sqrt[]{b} = \frac{a-b}{\sqrt[]{a}+\sqrt[]{b}} \)

微积分部分

极限部分

常用结论

x>0 \,时,\; \frac{x}{1+x} < \ln(1+x) < xx\in(0,\frac{\pi}{2})\,时,\;\sin{x}0且a\neq1)(\tan{x})'=(\sec{x})^2(\cot{x})'=-(\csc{x})^2(\sec{x})'=\sec{x}\tan{x}(\csc{x})'=- \, \csc{x} \, \cot{x}(\arcsin{x})'=\frac{1}{\sqrt[]{1-x^2}}(\arccos{x})'=\frac{-1}{\sqrt[]{1-x^2}}(\arctan{x})'=\frac{1}{1+x^2}[\, \ln {(x + \sqrt[]{1+x^2})} \,]' = \frac{1}{\sqrt[]{1+x^2}}

\(n阶导数公式\;\) (uv)^{(n)} = \sum_{k=0}^{n}C_n^ku^{(n-k)}v^{(k)}, \; 其中 \; C_n^k = \frac{n!}{k!(m-k)!}

(e^{ax})^{(n)} = a^n(e^{ax})\sin(ax)^{(n)} = a^n\sin(\frac{n\pi}{2} + ax)\cos(ax)^{(n)} = a^n\cos(\frac{n\pi}{2} + ax){\ln (1+x)}^{(n)}= \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}{ (1+x^{\alpha}) }^{(n)} = \alpha(\alpha - 1) \cdots (\alpha - n + 1)(1+x)^{\alpha - n}{ \frac{1}{x+a} }^{(n)} = (-1)^n \frac{n!}{(x+a)^(n+1)}

曲率计算公式

曲率 \quad k = \frac{\mid y^{''} \mid}{((1+(y^{'})^2)^{3/2} }曲率半径\quad R = \frac{1}{k}

基本积分公式

\int x^a\,dx = \frac{1}{a+1}x^{a+1} + C (a \neq -1)\int a^x\,dx=\frac{a^x}{\ln a} + C(a > 0, a \neq 1)\int \tan{x}\,dx = -\ln{\mid \, \cos{x} \, \mid}+C\int \sec{x}\,dx = \ln{\mid \, \sec{x}+\tan{x} \, \mid}+C\int \cot{x}\,dx = \ln{\mid \, \sin{x} \, \mid}+C\int \csc^2{x}\,dx = -\cot{x}+C\int \frac{1}{a^2+x^2}\,dx=\frac{1}{a}{\arctan\frac{x}{a}}+C(a \neq 0)\int \frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin \frac{x}{a}+C\int \frac{1}{a^2-x^2}\,dx=\frac{1}{2a}\ln \mid \, \frac{x-a}{x+a}+C \, \mid(a \neq 0)\int \frac{1}{\sqrt{x^2 \pm a^2}}\,dx=\ln{\mid \, x+\sqrt{x^2 \pm a^2}+C \, \mid}\int \, \sqrt[]{a^2-x^2} \, dx = \frac{x}{2} \sqrt[]{a^2-x^2} + \frac{a^2}{2} \arcsin \frac{x}{a}\int \, xe^{ax}\, dx = \frac{1}{a^2}(ax-1)e^{ax}

定积分的应用

极坐标曲线\; r = r(\theta)\; 介于两射线\;\theta=\alpha\;和\;\theta=\beta\,(\,0