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常用初等函数的导数和积分

导数

(xμ)′=μxμ−1\displaystyle {\left( x^{\mu}\right)}^\prime=\mu x^{\mu- 1} (sin⁡x)′=cos⁡x\displaystyle {\left( \sin{x}\right)}^\prime= \cos{x} (cos⁡x)′=−sin⁡x\displaystyle {\left( \cos{x}\right)}^\prime=- \sin{x} (tan⁡x)′=sec⁡2x\displaystyle {\left( \tan{x}\right)}^\prime={\sec}^{2} x (cot⁡x)′=−csc⁡2x\displaystyle {\left( \cot{x}\right)}^\prime=-{\csc}^{2} x (sec⁡x)′=sec⁡xtan⁡x\displaystyle {\left( \sec{x}\right)}^\prime= \sec{x} \tan{x} (csc⁡x)′=−csc⁡xcot⁡x\displaystyle {\left( \csc{x}\right)}^\prime=- \csc{x} \cot{x} (ax)′=axln⁡a\displaystyle {\left( a^{x}\right)}^\prime= a^{x} \ln{a} (ex)′=ex\displaystyle {\left(\text{e}^{x}\right)}^\prime=\text{e}^{x} (log⁡ax)′=1xln⁡a\displaystyle {\left({\log}_{a} x\right)}^\prime=\frac{1}{x \ln{a} } (ln⁡x)′=1x\displaystyle {\left( \ln{x}\right)}^\prime=\frac{1}{x} (arcsin⁡x)′=11−x2\displaystyle {\left( \arcsin{x}\right)}^\prime=\frac{1}{\sqrt{1- x^{2} } } (arccos⁡x)′=−11−x2\displaystyle {\left( \arccos{x}\right)}^\prime=-\frac{1}{\sqrt{1- x^{2} } } (arctan⁡x)′=11+x2\displaystyle {\left( \arctan{x}\right)}^\prime=\frac{1}{1+ x^{2} } (arccotx)′=−11+x2\displaystyle {\left(\text{arccot} x\right)}^\prime=-\frac{1}{1+ x^{2} }

积分

∫kdx=kx+C\displaystyle \int k{\left.\text{d} x\right.}= k x+ C ∫xμdx=1μ+1xμ+1+C,(μ≠−1)\displaystyle \int x^{\mu}{\left.\text{d} x\right.}=\frac{1}{\mu+ 1} x^{\mu+ 1}+ C,{\left(\mu\ne- 1\right)} ∫1xdx=ln⁡∣x∣+C\displaystyle \int\frac{1}{x}{\left.\text{d} x\right.}={\ln}{\left| x\right|}+ C ∫axdx=1ln⁡aax+C\displaystyle \int a^{x}{\left.\text{d} x\right.}=\frac{1}{\ln{a} } a^{x}+ C ∫sin⁡xdx=−cos⁡x+C\displaystyle \int \sin{x}{\left.\text{d} x\right.}=- \cos{x}+ C ∫cos⁡dx=sin⁡x+C\displaystyle \int \cos{\left.\text{d} x\right.}= \sin{x}+ C ∫1sin⁡2xdx=∫csc⁡2xdx=−cot⁡x+C\displaystyle \int\frac{1}{ {\sin}^{2} x}{\left.\text{d} x\right.}=\int{\csc}^{2} x{\left.\text{d} x\right.}=- \cot{x}+ C ∫1cos⁡2xdx=∫sec⁡2xdx=tan⁡x+C\displaystyle \int\frac{1}{ {\cos}^{2} x}{\left.\text{d} x\right.}=\int{\sec}^{2} x{\left.\text{d} x\right.}= \tan{x}+ C ∫1a2−x2dx=arcsin⁡(xa)+C,(a>0)∫11−x2dx=arcsin⁡x+C\begin{aligned}\displaystyle \int\frac{1}{\sqrt{a^{2}- x^{2} } }{\left.\text{d} x\right.}&= \arcsin{\left(\frac{x}{a}\right)}+ C,{\left( a> 0\right)} \\ \displaystyle \int\frac{1}{\sqrt{1- x^{2} } }{\left.\text{d} x\right.}&= \arcsin{x}+ C\end{aligned} ∫1a2+x2dx=1aarctan⁡(xa)+C,(a>0)∫11+x2dx=arctan⁡x+C\begin{aligned}\displaystyle \int\frac{1}{a^{2}+ x^{2} }{\left.\text{d} x\right.}&=\frac{1}{a} \arctan{\left(\frac{x}{a}\right)}+ C,{\left( a> 0\right)} \\ \displaystyle \int\frac{1}{1+ x^{2} }{\left.\text{d} x\right.}&= \arctan{x}+ C\end{aligned} ∫1a2−x2dx=12aln⁡∣a+xa−x∣+C,(a>0)∫1x2−a2dx=12aln⁡∣x−ax+a∣+C,(a>0)\begin{aligned}\displaystyle \int\frac{1}{a^{2}- x^{2} }{\left.\text{d} x\right.}&=\frac{1}{2 a}{\ln}{\left|\frac{a+ x}{a- x}\right|}+ C,{\left( a> 0\right)} \\ \displaystyle \int\frac{1}{x^{2}- a^{2} }{\left.\text{d} x\right.}&=\frac{1}{2 a}{\ln}{\left|\frac{x- a}{x+ a}\right|}+ C,{\left( a> 0\right)}\end{aligned} ∫tan⁡xdx=−ln⁡∣cos⁡x∣+C\displaystyle \int \tan{x}{\left.\text{d} x\right.}=-{\ln}{\left| \cos{x}\right|}+ C ∫cot⁡xdx=ln⁡∣sin⁡x∣+C\displaystyle \int \cot{x}{\left.\text{d} x\right.}={\ln}{\left| \sin{x}\right|}+ C ∫csc⁡xdx=∫1sin⁡xdx=ln⁡∣csc⁡x−cot⁡x∣+C=ln⁡∣tan⁡(x2)∣+C\begin{aligned}\displaystyle \int \csc{x}{\left.\text{d} x\right.}=\int\frac{1}{\sin{x} }{\left.\text{d} x\right.}&={\ln}{\left| \csc{x}- \cot{x}\right|}+ C \\ \displaystyle &={\ln}{\left| \tan{\left(\frac{x}{2}\right)}\right|}+ C\end{aligned} ∫sec⁡xdx=∫1cos⁡xdx=ln⁡∣sec⁡x+tan⁡x∣+C=ln⁡∣tan⁡(x2+π4)∣+C\begin{aligned}\displaystyle \int \sec{x}{\left.\text{d} x\right.}=\int\frac{1}{\cos{x} }{\left.\text{d} x\right.}&={\ln}{\left| \sec{x}+ \tan{x}\right|}+ C \\ \displaystyle &={\ln}{\left| \tan{\left(\frac{x}{2}+\frac{\pi}{4}\right)}\right|}+ C\end{aligned} ∫1x2±a2dx=ln⁡∣x+x2±a2∣+C\displaystyle \int\frac{1}{\sqrt{x^{2}\pm a^{2} } }{\left.\text{d} x\right.}={\ln}{\left| x+\sqrt{x^{2}\pm a^{2} }\right|}+ C ∫a2−x2dx=a22arcsin⁡(xa)+x2a2−x2+C\displaystyle \int\sqrt{a^{2}- x^{2} }{\left.\text{d} x\right.}=\frac{a^{2} }{2} \arcsin{\left(\frac{x}{a}\right)}+\frac{x}{2}\sqrt{a^{2}- x^{2} }+ C ∫x2±a2dx=x2x2±a2±a22ln⁡∣x+x2±a2∣+C\displaystyle \int\sqrt{x^{2}\pm a^{2} }{\left.\text{d} x\right.}=\frac{x}{2}\sqrt{x^{2}\pm a^{2} }\pm\frac{a^{2} }{2}{\ln}{\left| x+\sqrt{x^{2}\pm a^{2} }\right|}+ C

表格积分法

例 1

∫x3e2xdx\displaystyle \int x^{3}\text{e}^{2 x}{\left.\text{d} x\right.}

public/math/table-integral.excalidraw.svg

∫x3e2xdx=12x3e2x−34x2e2x+34xe2x−38e2x+C=(12x3−34x2+34x−38)e2x+C\begin{aligned}\displaystyle \int x^{3}\text{e}^{2 x}{\left.\text{d} x\right.}&=\frac{1}{2} x^{3}\text{e}^{2 x}-\frac{3}{4} x^{2}\text{e}^{2 x}+\frac{3}{4} x\text{e}^{2 x}-\frac{3}{8}\text{e}^{2 x}+ C \\ \displaystyle &={\left(\frac{1}{2} x^{3}-\frac{3}{4} x^{2}+\frac{3}{4} x-\frac{3}{8}\right)}\text{e}^{2 x}+ C\end{aligned}

例 2

∫x2cos⁡2xdx\displaystyle \int x^{2} \cos{2} x{\left.\text{d} x\right.}

public/math/table-integral-2.excalidraw.svg

∫x2cos⁡2xdx=12xsin⁡2x+12xcos⁡2x−14sin⁡2x+C\displaystyle \int x^{2} \cos{2} x{\left.\text{d} x\right.}=\frac{1}{2} x \sin{2} x+\frac{1}{2} x \cos{2} x-\frac{1}{4} \sin{2} x+ C

例 3

∫e2xcos⁡xdx\displaystyle \int\text{e}^{2 x} \cos{x}{\left.\text{d} x\right.}

public/math/table-inte-3.excalidraw.svg

∫e2xcos⁡xdx=12e2xcos⁡x+14e2xsin⁡x−14∫e2xcos⁡xdx∫e2xcos⁡xdx=45(12e2xcos⁡x+14e2xsin⁡x)+C\begin{aligned}\displaystyle \int\text{e}^{2 x} \cos{x}{\left.\text{d} x\right.}&=\frac{1}{2}\text{e}^{2 x} \cos{x}+\frac{1}{4}\text{e}^{2 x} \sin{x}-\frac{1}{4}\int\text{e}^{2 x} \cos{x}{\left.\text{d} x\right.} \\ \displaystyle \int\text{e}^{2 x} \cos{x}{\left.\text{d} x\right.}&=\frac{4}{5}{\left(\frac{1}{2}\text{e}^{2 x} \cos{x}+\frac{1}{4}\text{e}^{2 x} \sin{x}\right)}+ C\end{aligned}

变限积分

  • f(x)\displaystyle f{\left( x\right)} 为偶函数，则 F(x)=∫0xf(t)dt\displaystyle F{\left( x\right)}=\int_{0}^{x} f{\left( t\right)}\text{d} t 为奇函数，注意，必须满足 F(0)=0\displaystyle F{\left( 0\right)}= 0
  • f(x)\displaystyle f{\left( x\right)} 为奇函数，则 F(x)=∫axf(t)dt\displaystyle F{\left( x\right)}=\int_{a}^{x} f{\left( t\right)}\text{d} t 为偶函数，没有上面的限制，因此下限可以是 a\displaystyle a

多元隐函数求导

如果用“隐函数存在定理”求，即通过计算 Fx′,Fy′,Fz′\displaystyle F_{x}^\prime, F_{y}^\prime, F_{z}^\prime，并且用 ∂z∂x=−∂Fx′∂Fz′\displaystyle \frac{\partial z}{\partial x}=-\frac{\partial F_{x}^\prime}{\partial F_{z}^\prime} 计算偏导数，一定要注意，F(x,y,z)\displaystyle{ F \left( x , y , z \right) } 关于求 x\displaystyle{ x } 偏导的时候，y\displaystyle{ y } 和 z\displaystyle{ z } 需要当作常数，F(x,y,z)\displaystyle{ F \left( x , y , z \right) } 关于 y\displaystyle{ y } 和 z\displaystyle{ z } 求偏导的时候也是。

例：z=z(x,y)\displaystyle{ z = z \left( x , y \right) } 是由 x2+y2+z2−4z=0\displaystyle{ x ^{ 2 } + y ^{ 2 } + z ^{ 2 } - 4 z = 0 } 确定的函数，求 ∂z∂x\displaystyle \frac{\partial z}{\partial x}

如果设 F(x,y,z)=x2+y2+z2−4z\displaystyle{ F \left( x , y , z \right) = x ^{ 2 } + y ^{ 2 } + z ^{ 2 } - 4 z }，那么

Fx′=2xFz′=2z−4\begin{aligned}\displaystyle F_{x}^\prime&= 2 x \\ \displaystyle F_{z}^\prime&= 2 z- 4\end{aligned} ∂z∂x=−Fx′Fz′=−xz−2\displaystyle \frac{\partial z}{\partial x}=-\frac{F_{x}^\prime}{F_{z}^\prime}=-\frac{x}{z- 2}

不要把第一个式子中的 z\displaystyle z 当作 x\displaystyle{ x } 的函数，而应该将 z\displaystyle{ z } 看作是常数。