基本初等函数的导数

这里列出六种基本初等函数的导数公式，并给出其证明。

一阶导数

常数函数

若 $f(x) = c, c \in \mathbb{R}$，则 $f'(x) = 0.$

关于这个定理/命题的证明，单击这里以显示/折叠

$\begin{align} f'(x_0) & = \lim _{x \to x_0} \dfrac{f(x) - f(x_0)}{x - x_0} \\ & = \lim _{x \to x_0} \dfrac{c - c}{x - x_0} \\ & = 0. \end{align}$

幂函数

若 $f(x) = x^\alpha, \alpha \in \mathbb{R}, x > 0$，则 $f'(x) = \alpha x^{\alpha-1}.$

关于这个定理/命题的证明，单击这里以显示/折叠

$\begin{align} f'(x_0) & = \lim _{x \to x_0} \dfrac{f(x) - f(x_0)}{x - x_0} \\ & = \lim _{x \to x_0} x_0^\alpha \dfrac{ (\frac{x}{x_0})^\alpha - 1 }{x - x_0} \\ & = \lim _{x \to x_0} x_0^\alpha \dfrac{ \alpha (\frac{x}{x_0} - 1) }{x - x_0} \\ & = \alpha x_0^{\alpha-1}. \end{align}$

指数函数

若 $f(x) = a^x, a \in \mathbb{R^{+}}$，则 $f'(x) = a^x \ln a.$

关于这个定理/命题的证明，单击这里以显示/折叠

$\begin{align} f'(x_0) & = \lim _{x \to x_0} \dfrac{f(x) - f(x_0)}{x - x_0} \\ & = \lim _{x \to x_0} \dfrac{ a^x - a^{x_0} }{x - x_0} \\ & = \lim _{x \to x_0} a^{x_0} \dfrac{ a^{x - x_0} - 1 }{x - x_0} \\ & = \lim _{x \to x_0} a^{x_0} \dfrac{ (x - x_0) \ln a }{x - x_0} \\ & = a^{x_0} \ln a. \end{align}$

对数函数

若 $f(x) = \log _a x, a \in (0, 1) \cup (1, +\infty)$，则 $f'(x) = \dfrac{1}{x \ln a}.$

关于这个定理/命题的证明，单击这里以显示/折叠

$\begin{align} f'(x_0) & = \lim _{x \to x_0} \dfrac{f(x) - f(x_0)}{x - x_0} \\ & = \lim _{x \to x_0} \dfrac{ \log _a x - \log _a x_0 }{x - x_0} \\ & = \lim _{x \to x_0} \dfrac{ \ln x - \ln x_0 }{(x - x_0) \ln a} \\ & = \lim _{x \to x_0} \dfrac{ \ln ( 1 + \frac{x}{x_0} - 1 )}{(x - x_0) \ln a} \\ & = \lim _{x \to x_0} \dfrac{ \frac{x}{x_0} - 1 }{(x - x_0) \ln a} \\ & = \dfrac{1}{x_0 \ln a}. \end{align}$

三角函数

$\begin{array}{cclccl} (\sin x)' & = & \cos x & (\cos x)' & = & - \sin x \\ (\tan x)' & = & \sec^2 x & (\cot x)' & = & - \csc^2 x \\ (\sec x)' & = & \sec x \tan x & (\csc x)' & = & - \csc x \cot x \end{array}$

关于这个定理/命题的证明，单击这里以显示/折叠

$\begin{align} (\sin x_0)' & = \lim _{x \to x_0} \dfrac{ \sin x - \sin x_0 }{x - x_0} \\ & = \lim _{x \to x_0} \dfrac{ 2 \sin \frac{x-x_0}{2} \cos \frac{x+x_0}{2} }{x - x_0} \\ & = \lim _{x \to x_0} \dfrac{ 2 \frac{x-x_0}{2} \cos \frac{x_0+x_0}{2} }{x - x_0} \\ & = \cos x_0. \\ (\cos x_0)' & = \lim _{x \to x_0} \dfrac{ \cos x - \cos x_0 }{x - x_0} \\ & = \lim _{x \to x_0} \dfrac{ -2 \sin \frac{x-x_0}{2} \sin \frac{x+x_0}{2} }{x - x_0} \\ & = \lim _{x \to x_0} \dfrac{ -2 \frac{x-x_0}{2} \sin \frac{x_0+x_0}{2} }{x - x_0} \\ & = - \sin x_0. \\ (\tan x)' & = \left( \dfrac{\sin x}{\cos x} \right)' \\ & = \dfrac{\cos x \cos x - (-\sin x) \sin x}{\cos^2 x} \\ & = \dfrac{1}{\cos^2 x}. \\ (\cot x)' & = \left( \dfrac{\cos x}{\sin x} \right)' \\ & = \dfrac{-\sin x \sin x - \cos x \cos x}{\sin^2 x} \\ & = - \dfrac{1}{\sin^2 x}. \\ (\sec x)' & = \left( \dfrac{1}{\cos x} \right)' = \dfrac{\sin x}{\cos^2 x} = \sec x \tan x. \\ (\csc x)' & = \left( \dfrac{1}{\sin x} \right)' = \dfrac{- \cos x}{\sin^2 x} = - \csc x \cot x. \end{align}$

反三角函数

${\begin{array}{cclccl}(\arcsin x)'&=&{\dfrac {1}{\sqrt {1-x^{2}}}}&(\arccos x)'&=&-{\dfrac {1}{\sqrt {1-x^{2}}}}\\(\arctan x)'&=&{\dfrac {1}{1+x^{2}}}&(\operatorname {arccot} x)'&=&-{\dfrac {1}{1+x^{2}}}\\(\operatorname {arcsec} x)'&=&{\dfrac {1}{|x|{\sqrt {x^{2}-1}}}}&(\operatorname {arccsc} x)'&=&-{\dfrac {1}{|x|{\sqrt {x^{2}-1}}}}\end{array}}$

关于这个定理/命题的证明，单击这里以显示/折叠

${\begin{aligned}(\arcsin x)'&={\dfrac {1}{(\sin y)'|_{y=\arcsin x}}}={\dfrac {1}{\cos(\arcsin x)}}={\dfrac {1}{\sqrt {1-x^{2}}}}.\\(\arccos x)'&={\dfrac {1}{(\cos y)'|_{y=\arccos x}}}={\dfrac {1}{-\sin(\arccos x)}}=-{\dfrac {1}{\sqrt {1-x^{2}}}}.\\(\arctan x)'&={\dfrac {1}{(\tan y)'|_{y=\arctan x}}}\\&={\dfrac {1}{\sec ^{2}(\arctan x)}}={\dfrac {1}{1+\tan ^{2}(\arctan x)}}={\dfrac {1}{1+x^{2}}}.\\(\operatorname {arccot} x)'&={\dfrac {1}{(\cot y)'|_{y=\operatorname {arccot} x}}}\\&={\dfrac {1}{-\csc ^{2}(\operatorname {arccot} x)}}=-{\dfrac {1}{1+\cot ^{2}(\operatorname {arccot} x)}}=-{\dfrac {1}{1+x^{2}}}.\\(\operatorname {arcsec} x)'&={\dfrac {1}{(\sec y)'|_{y=\operatorname {arcsec} x}}}={\dfrac {1}{\sec(\operatorname {arcsec} x)\tan(\operatorname {arcsec} x)}}={\dfrac {1}{|x|{\sqrt {x^{2}-1}}}}.\\(\operatorname {arccsc} x)'&={\dfrac {1}{(\csc y)'|_{y=\operatorname {arccsc} x}}}={\dfrac {1}{-\csc(\operatorname {arccsc} x)\cot(\operatorname {arccsc} x)}}=-{\dfrac {1}{|x|{\sqrt {x^{2}-1}}}}.\end{aligned}}$

相关章节

• 上一节：导数
• 下一节：求导法则

参考资料

1. 欧阳光中, 朱学炎, 金福临, 陈传璋, 《数学分析》, 高等教育出版社, 北京, 2018-08, ISBN 978-7-0404-9718-2.