公式答案

导数

\((C){’} =\) \(0\)

\((X^{\mu}){’} = \mu{x}^{\mu-1}\)

\((\sqrt{x}){’} =\frac{1}{2\sqrt{x}}\)

\((\frac{1}{x}){’} = -\frac{1}{x^2}\)

\((e^{x}){’} =e^{x}\)

\((a^{x}){’} =a^{x}\ln{a}\)

\((\ln{x}){’} =\frac{1}{x}\)

\((\log_{a}{x}){’} =\frac{1}{x\ln{a}}\)

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

\((\sin{x}){’} =\cos{x}\)

\((\tan{x}){’} =\sec^2{x}\)

\((\sec{x}){’} =\sec{x}\tan{x}\)

\((\cos{x}){’} =-\sin{x}\)

\((\cot{x}){’} =-\csc^2{x}\)

\((\csc{x}){’} =-csc{x}cot{x}\)

\((\arcsin{x}){’} =\frac{1}{\sqrt{1-x^2}}\)

\((\arccos{x}){’} =-\frac{1}{\sqrt{1-x^2}}\)

\((\arctan{x}){’} =\frac{1}{1+x^2}\)

\((\text{arccot}\ x){’} =-\frac{1}{1+x^2}\)

三角函数

\(y=\sin x\)

\(y=\cos x\)

\(y=\tan x\)

\(y=\cot x\)

\(y=\sec x\)

\(y=\csc x\)

上述函数图像

诱导公式

奇变偶不变，符号看象限。

\(\sin(α+2kπ)=\sin α\)
\(\cos(α+2kπ)=\cos α\)
\(\tan(α+2kπ)=\tanα\)
\(\cot(α+2kπ)=\cotα\)

\(\sin(-α)=-\sinα\)
\(\cos(-α)=cos α\)
\(\tan(-α)=-\tan α\)
\(\cot(-α)=-cot α\)

\(\sin(α+π)=-\sin α\)
\(\cos(α+π)=-\cos α\)
\(\tan(α+π)=\tan α\)
\(\cot(α+π)=cot α\)

\(\sin(π-α)=\sin α\)
\(\cos(π-α)=-\cos α\)
\(\tan(π-α)=-\tan α\)
\(\cot(π-α)=-\cot α\)

\(\sin(α+\frac{\pi}{2})=\cos α\)
\(\cos(α+\frac{\pi}{2})=-\sin α\)
\(\tan(α+\frac{\pi}{2})=-\cot α\)
\(\cot(α+\frac{\pi}{2})=-\tan α\)

\(\sin(\frac{\pi}{2}-α)=\cos α\)
\(\cos(\frac{\pi}{2}-α) =\sin α\)
\(\tan(\frac{\pi}{2}-α)=\cot α\)
\(\cot(\frac{\pi}{2}-α)=\tan α\)

两角和与差

\(\cos(\alpha + \beta) =\cosα\cosβ-\sinα\sinβ\)

\(\cos{(\alpha - \beta)} =\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}\)

\(\sin{(\alpha + \beta)} =\sin{\alpha}\cos{\beta}+cos{\alpha}\sin{\beta}\)

\(\sin{(\alpha - \beta)} = \sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}\)

\(\tan{(\alpha + \beta)} = \frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}\)

\(\tan{(\alpha - \beta)} = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}\)

和差化积

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

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

\(\cos\alpha + \cos\beta = 2\cos(\frac{\alpha + \beta}{2})\cos(\frac{\alpha - \beta}{2})\)

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

积化和差

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

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

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

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

二倍角公式

\(\sin{2\alpha} = 2\sin\alpha\cos\alpha = \frac{2}{\tan\alpha + \cot\alpha}\)

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

\(\tan{2\alpha} = \frac{2\tan\alpha}{1 - \tan^2{\alpha}}\)

\(\cot{2\alpha} = \frac{\cot^2{\alpha} - 1}{2\cot\alpha}\)

\(\sec{2\alpha} = \frac{\sec^2{\alpha}}{1 - \tan^2{\alpha}}\)

\(\csc{2\alpha} = \frac{1}{2\sin\alpha\cos\alpha} = \frac{1}{2}\sec\alpha\csc\alpha\)

降幂公式

\(\sin^2\alpha = \frac{1 - cos2\alpha}{2}\)

\(\cos^2\alpha = \frac{1 +\cos2\alpha}{2}\)

\(\tan^2\alpha = \frac{1 - \cos2\alpha}{1 + \cos2\alpha}\)

辅助角公式

\(a\sin\alpha + b\cos\alpha = \sqrt{a^2 +b^2}\sin(\alpha + \varphi)\)

\(其中\ \varphi\ 满足\ \cos\varphi = \frac{a}{\sqrt{a^2 + b^2}}，\sin\varphi = \frac{b}{\sqrt{a^2 + b^2}}\)

万能公式

\(\sin\alpha = \frac{2\tan\frac{a}{2}}{1 + \tan^2\frac{a}{2}}\)

\(\cos\alpha = \frac{1 - \tan^2\frac{a}{2}}{1 + \tan^2\frac{a}{2}}\)

\(\tan\alpha = \frac{2\tan\frac{a}{2}}{1-\tan^2\frac{a}{2}}\)

反三角函数

\(\arcsin{x}\)

\(\arccos{x}\)

\(\arctan{x}\)

\({\mathrm{arccot} {x}}\)

上述反三角函数的定义域、值域、图像：

积分公式

\(\int{x^{\mu}dx} = \frac{1}{\mu + 1}x^{\mu+1} + C\)

\(\int{\frac{1}{\sqrt{x}}dx} = 2\sqrt{x} + C\)

\(\int{e^x dx} = e^x + C\)

\(\int{a^x dx} = \frac{a^x}{\ln{a}} + C\)

\(\int{\frac{1}{x}dx} = \ln|x| + C\)

\(\int{\sin{x}dx} = -\cos{x} + C\)

\(\int{\cos{x}dx} = \sin{x} + C\)

\(\int{\tan{x}dx} = \int{\frac{\sin{x}}{\cos{x}}dx} = -\int{\frac{d\cos{x}}{\cos{x}}} = -\ln|\cos{x}| + C\)

\(\int{\cot{x}dx} = \int{\frac{\cos{x}}{\sin{x}}dx} = \int{\frac{d\sin{x}}{\sin{x}}} = \ln|\sin{x}| + C\)

\(\int{\sec^2{x}dx} = \int{\frac{dx}{\cos^2x}} = \tan{x} + C\)

\(\int{\csc^2{x}dx} = \int{\frac{dx}{\sin^2x}} = -\cot{x} + C\)

\(\int{\frac{dx}{\sqrt{1-x^2}}} = \arcsin{x} + C\)

\(\int{\frac{dx}{1+x^2}} = \arctan{x} + C\)

\(\int{\sec{x}dx} = \ln|\sec{x} + \tan{x}| + C\)

\(\int{\csc{x}dx} = \ln|\csc{x} - \cot{x}| + C\)

\(\int{\frac{1}{\sqrt{a^2 - x^2}}dx} = \arcsin{\frac{x}{a}} + C\)

\(\int{\frac{1}{a^2 + x^2}dx} = \frac{1}{a}\arctan{\frac{x}{a}} + C\)

\(\int{\frac{1}{\sqrt{x^2 \pm a^2}}dx} = \ln|x + \sqrt{x^2 \pm a^2}| + C\)

\(\int{\frac{1}{1 + e^x}dx} = x - \ln(1 + e^x) + C\)

\(\int{\frac{1}{a^2 - x^2}dx} = \frac{1}{2a}\ln|{\frac{a+x}{a-x}}| + C\)

\(\int{\sqrt{x^2 + a^2}dx} = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln{(x+\sqrt{x^2 + a^2}) + C}\)

\(\int{\sqrt{x^2 - a^2}dx} = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln|{x+\sqrt{x^2 - a^2}| + C}\)

\(\int{\sqrt{a^2 - x^2}dx} = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin{\frac{x}{a}} + C\)

\(\int{\sec^3{x}dx} = \frac{1}{2}(\sec{x}\tan{x}+\ln|\sec{x}+\tan{x}|)\)

麦克劳林公式

\(\sin x = x - \frac{1}{6}x^3 + o(x^3) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 + o(x^5) = \sum\limits_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\)

\(\arcsin x = x + \frac{1}{6}x^3 + o(x^3) = \sum\limits_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}\cdot\frac{x^{2n+1}}{2n+1}\)

\(\tan x = x + \frac{1}{3}x^3 + o(x^3) = \sum\limits_{n=1}^{\infty}\frac{(2^{2n}-1)2^{2n}B_n}{(2n)!}x^{2n-1}，\{B_n\}为伯努利数，即B_1=\frac{1}{6},B_2=\frac{1}{30},\cdot\cdot\cdot\)

\(\arctan x = x - \frac{1}{3}x^3 + o(x^3)\)

\(\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) + ...+(-1)^{n}\frac{x^{2n}}{（2n）!}= \sum\limits_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n)!}\)

\(\ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + o(x^3) = \sum\limits_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}\)

\(a^x = 1+ x\ln{a} + \frac{1}{2}{(x\ln{a})}^2 + \frac{1}{6}{(x\ln{a})}^3 + o(x^3) = \sum\limits_{n=0}^{\infty}\frac{{(x\ln{a})}^n}{n!}\)

\(e^x = 1+ x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + o(x^3) = \sum\limits_{n=0}^{\infty}\frac{x^n}{n!}\)

\((1+x)^a = 1 + ax + \frac{a(a-1)}{2}x^2 + o(x^2) =1 + \sum\limits_{n=1}^{\infty}\frac{a(a-1)\cdot\cdot\cdot(a-n+1)}{n!}x^n\)

\(\frac{1}{1+x} = 1 - x + x^2 - x^3 + o(x^3) = \sum\limits_{n=0}^{\infty}{(-1)^n}{x^n}\)

\(\frac{1}{1-x} = 1 + x + x^2 + x^3 + o(x^3) = \sum\limits_{n=0}^{\infty}x^n\)

\(\sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{2 \cdot 4}x^2 + \frac{1 \cdot 3}{2 \cdot 4 \cdot 6}x^3 - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8}x^4 + o(x^4)\)

\(\frac{1}{\sqrt {1+x}} = 1 - \frac{1}{2}x + \frac{1}{2 \cdot 4}x^2 - \frac{1 \cdot 3}{2 \cdot 4 \cdot 6}x^3 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8}x^4 + o(x^4)\)

重要差函数

\(x - \sin{x} \sim \frac{1}{6}x^3\)

\(\arcsin{x} - x \sim \frac{1}{6}x^3\)

\(1 - \cos{x} \sim \frac{1}{2}x^2\)

\(x - \ln{(1+x)} \sim \frac{1}{2}x^2\)

\(\tan{x} - x \sim \frac{1}{3}x^3\)

\(e^x - 1 \sim x\)

\(e^x - 1 - x \sim \frac{1}{2}x^2\)

\(\tan{x} - \sin{x} \sim \frac{1}{2}x^3\)

推广：\(\tan{f(x)} - \sin{f(x)} \sim \frac{1}{2}f^3(x)\)

\(\arctan{x} - \arcsin{x} \sim -\frac{1}{2}x^3\)

注意：\(f(x) \pm g(x) = cx^k + o(x^k)\) ，而 \(f(x) \pm g(x) \sim cx^k\) 。

高阶导数

\([\sin(ax+b)]^{(n)}={a^n}sin(ax+b+\frac{n\pi}{2})\)

\([\cos(ax+b)]^{(n)}={a^n}cos(ax+b+\frac{n\pi}{2})\)

\([\ln(ax+b)]^{(n)}=(-1)^{n-1}a^{n}\frac{(n-1)!}{(ax+b)^n}\)

\((\frac{1}{ax+b})^{(n)}=(-1)^n a^n \frac{n!}{(ax+b)^{n+1}}\)

\((e^{ax+b})^{(n)}= a^n e^{ax+b}\)