等价无穷小
前提：\(x\rightarrow 0\)
\(e^x-1\sim x\qquad a^x-1\sim x\ln{a}\)
\(\ln(1+x)\sim x\qquad\log_a(1+x)\sim \dfrac{x}{\ln{a}}\)
\(x-\ln(1+x)\sim\dfrac{1}{2}x^2\)
\((1+x)^a-1\sim ax,a\neq 0\qquad (1+x)^n-1\sim nx\)
\(拓展:当A(x)\rightarrow 0且A(x)B(x)\rightarrow 0,[1+A(x)]^{B(x)}-1\sim A(x)B(x)\)
\(\sin{x}\sim\arcsin{x}\sim\tan{x}\sim\arctan{x}\sim x\)
\([x-\sin{x}]\sim[\arcsin{x}-x]\sim\dfrac{1}{3!}x^3\)
\([\tan{x}-x]\sim[x-\arctan{x}]\sim\dfrac{1}{3}x^3\)
\(1-\cos{x}\sim\dfrac{1}{2}x^2\)
\(拓展:1-\cos^a{x}\sim\dfrac{a}{2}x^2\)
导数+积分
基本公式
\((C)^{\prime}=0\)
\(\begin{aligned}(x^a)^{\prime}=ax^{a-1}\qquad \int x^adx=\frac{1}{a+1}x^{a+1}+C\end{aligned}\)
\((e^x)^{\prime}=e^x\qquad(a^x)^{\prime}=a^x\ln{a}\)
\((\ln{|x|})^{\prime}=\dfrac{1}{x}\qquad (\log_{a}{x})^{\prime}=\dfrac{1}{x\ln{a}}\)
\(|x|^{\prime}=\dfrac{x}{|x|}\)
\((\sin{x})^{\prime}=\cos{x}\qquad(\cos{x})^{\prime}=-\sin{x}\)
\((\tan{x})^{\prime}=\sec^2{x}\qquad(\cot{x})^{\prime}=-\csc^2{x}\)
\((\sec{x})^{\prime} = \sec{x} \tan{x} \qquad (\csc{x})^{\prime} = -\csc{x} \cot{x}\)
\(\begin{aligned} \int \sec{x}dx = \ln{|\sec{x} + \tan{x}|} + C \qquad \int \csc{x}dx = \ln{|\csc{x} - \cot{x}|} + C \end{aligned}\)
结合三角恒等变换公式
\(\begin{aligned}求\int\sin^2{x}dx,用\cos{2x}=1-2\sin^2{x}\end{aligned}\)
\(\begin{aligned}求\int\cos^2{x}dx,用\cos{2x}=2\cos^2{x}-1\end{aligned}\)
\(\begin{aligned}求\int\tan^2{x}dx,用\tan^2{x}+1=\sec^2{x}\end{aligned}\)
\(\begin{aligned}求\int\cot^2{x}dx,用\cot^2{x}+1=\csc^2{x}\end{aligned}\)
\(\begin{aligned} (\arcsin{x})^{\prime} = -(\arccos{x})^{\prime} = \dfrac{1}{\sqrt{1 - x^2}} \qquad \int \dfrac{1}{\sqrt{a^2 - x^2}}dx= \arcsin{\dfrac{x}{a}} + C \end{aligned}\)
\(\begin{aligned} (\arctan{x})^{\prime} = -(\mathrm{arccot}\,{x})^{\prime} = \dfrac{1}{1 + x^2} \qquad \int \dfrac{1}{a^2 + x^2}dx= \dfrac{1}{a} \arctan{\dfrac{x}{a}} + C \end{aligned}\)
\(\begin{aligned} \int \dfrac{1}{\sqrt{x^2 \pm a^2}}dx= \ln{|x + \sqrt{x^2 \pm a^2}|} + C \end{aligned}\)
\(\begin{aligned} \int \dfrac{1}{a^2 - x^2}dx= \dfrac{1}{2a} \ln{\left| \dfrac{a + x}{a - x} \right|} + C \end{aligned}\)
常用快速公式
\(\left(\dfrac{1}{x}\right)^{\prime}=-\dfrac{1}{x^2}\)
\((\sqrt{x})^{\prime}=\dfrac{1}{2\sqrt{x}}\)
\((\sin^2{x})^{\prime}=-(\cos^2{x})^{\prime}=\sin{2x}\)
\(\begin{aligned}\int_0^\frac{\pi}{2}\sin{x}dx=\int_0^\frac{\pi}{2}\cos{x}dx=1\end{aligned}\)
\(\begin{aligned}\Gamma(n+1)=\int_{0}^{+\infty}x^{n}e^{-x}dx=n!\end{aligned}\)
\(\begin{aligned}\int_0^1\ln{x}dx=-1\end{aligned}\)
高阶导数
\(\begin{aligned}莱布尼兹公式:(uv)^{(n)} = \sum_{k=0}^{n} C_{n}^{k} u^{(n-k)} v^{(k)},其中u^{(0)}=u\end{aligned}\)
\(\bigg[\sin(ax+b)\bigg]^{(n)}=a^n\sin(ax+b+\dfrac{\pi}{2}n)\)
\(\bigg[\cos(ax+b)\bigg]^{(n)}=a^n\cos(ax+b+\dfrac{\pi}{2}n)\)
\(\bigg(\dfrac{1}{ax+b}\bigg)^{(n)}=a^n\dfrac{(-1)^nn!}{(ax+b)^{n+1}}\)
\((x^n)^{(n)}=n!\)
泰勒+麦克劳林展开式
\(奇函数展开后只有奇数次幂(\sin{x},\tan{x},\arcsin{x},\arctan{x},\dfrac{e^x-e^{-x}}{2})\)
\(偶函数展开后只有偶数次幂(\cos{x})\)
\(\begin{aligned}几何级数:\dfrac{1}{1-x}=\sum\limits_{n=0}^{\infty}x^n=1+x+x^2+\cdots,x\in(-1,1)\end{aligned}\)
\(\begin{aligned}换元得:\dfrac{1}{1+x}=\sum\limits_{n=0}^{\infty}(-1)^nx^n=1-x+x^2-\cdots,x\in(-1,1)\end{aligned}\)
\(\begin{aligned}二项式级数:(1+x)^a=1+ax+\dfrac{a(a-1)}{2!}x^2+\cdots+\dfrac{\overbrace{a(a-1)\cdots(a-n+1)}^{n项}}{n!}x^n+\cdots,x\in(-1,1)\end{aligned}\)
\(\begin{aligned} e^x = \sum\limits_{n=0}^{\infty} {\dfrac{x^n}{n!}} = 1 + x + \dfrac{x^2}{2!} + \cdots,x \in (-\infty,+\infty) \end{aligned}\)
\(\begin{aligned} \ln{(1+x)} = \sum\limits_{n=1}^{\infty} (-1)^{n+1} \dfrac{x^{n}}{n} = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots,x \in (-1,1] \end{aligned}\)
\(\begin{aligned}换元得: -\ln{(1-x)} = \sum\limits_{n=1}^{\infty} \dfrac{x^{n}}{n} = x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots,x \in [-1,1) \end{aligned}\)
\(\begin{aligned} \sin{x} = \sum\limits_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!} = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots,x \in (-\infty,+\infty) \end{aligned}\)
\(\begin{aligned} 求导得:\cos{x} = \sum\limits_{n=0}^{\infty} (-1)^n \dfrac{x^{2n}}{(2n)!} = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots,x \in (-\infty,+\infty) \end{aligned}\)
\(\begin{aligned} 拓展:\dfrac{e^x - e^{-x}}{2} = \sum\limits_{n=0}^{\infty} {\dfrac{x^{2n+1}}{(2n+1)!}},x \in (-\infty,+\infty) \end{aligned}\)
\(\begin{aligned} \arctan{x} = \sum\limits_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{2n+1} = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots,x \in [-1,1] \end{aligned}\)
\(\tan{x}=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\cdots\)
\(\arcsin{x} = x + \left(\dfrac{1}{2}\right) \dfrac{x^3}{3} + \left(\dfrac{1 \cdot 3}{2 \cdot 4}\right) \dfrac{x^5}{5} + \cdots\)
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