基本初等函数的微分公式与微分运算法则

 

基本初等函数的微分公式

 

导数公式	微分公式
\[\left( {{x^\mu }} \right)' = \mu {x^{\mu - 1}}\]	\[{\text{d}}\left( {{x^\mu }} \right) = \mu {x^{\mu - 1}}{\text{d}}x\]
\[\left( {\sin x} \right){'} = \cos x\]	\[{\text{d}}\left( {\sin x} \right) = \cos x{\text{d}}x\]
\[\left( {\cos x} \right)' = - \sin x\]	\[{\text{d}}\left( {\cos x} \right) = - \sin x{\text{d}}x\]
\[(\tan x)' = {\text{se}}{{\text{c}}^2}x\]	\[{\text{d}}\left( {\tan x} \right) = {\text{se}}{{\text{c}}^2}x{\text{d}}x\]
\[\left( {\cot x} \right)' = - {\text{cs}}{{\text{c}}^2}x\]	\[{\text{d}}\left( {\cot x} \right) = - {\text{cs}}{{\text{c}}^2}x{\text{d}}x\]
$$\begin{array}{l} \left( {\sec x} \right)' = \\ \sec x\tan x \end{array}$$	$$\begin{array}{l} {\rm{d}}\left( {\sec x} \right) = \\ \sec x\tan x{\rm{d}}x \end{array}$$
$$\begin{array}{l} \left( {\csc x} \right)' = \\ - \csc x\cot x \end{array}$$	$$\begin{array}{l} {\rm{d}}\left( {\csc x} \right) = \\ - \csc x\cot x{\rm{d}}x \end{array}$$
\[\left( {{a^x}} \right)' = {a^x}\ln a\]	\[{\text{d}}\left( {{a^x}} \right) = {a^x}\ln a{\text{d}}x\]
\[\left( {{{\text{e}}^x}} \right)' = {{\text{e}}^x}\]	\[{\text{d}}\left( {{{\text{e}}^x}} \right) = {{\text{e}}^x}{\text{d}}x\]
\[\left( {{{\log }_a}x} \right){'} = \frac{1}{{x\ln a}}\]	\[{\text{d}}\left( {{{\log }_a}x} \right) = \frac{1}{{x\ln a}}{\text{d}}x\]
\[\left( {\ln x} \right){'} = \frac{1}{x}\]	\[{\text{d}}\left( {\ln x} \right) = \frac{1}{x}{\text{d}}x\]
$$\begin{array}{l} \left( {\arcsin x} \right)' = \\ \frac{1}{{\sqrt {1 - {x^2}} }} \end{array}$$	$$\begin{array}{l} {\rm{d}}\left( {\arcsin x} \right) = \\ \frac{1}{{\sqrt {1 - {x^2}} }}{\rm{d}}x \end{array}$$
$$\begin{array}{l} \left( {\arccos x} \right)' = \\ - \frac{1}{{\sqrt {1 - {x^2}} }} \end{array}$$	$$\begin{array}{l} {\rm{d}}\left( {\arccos x} \right) = \\ - \frac{1}{{\sqrt {1 - {x^2}} }}{\rm{d}}x \end{array}$$
$$\begin{array}{l} \left( {\arctan x} \right)' = \\ \frac{1}{{1 + {x^2}}} \end{array}$$	$$\begin{array}{l} {\rm{d}}\left( {\arctan x} \right) = \\ \frac{1}{{1 + {x^2}}}{\rm{d}}x \end{array}$$
$$\begin{array}{l} \left( {{\rm{arccot}}x} \right)' = \\ - \frac{1}{{1 + {x^2}}} \end{array}$$	$$\begin{array}{l} {\rm{d}}\left( {{\rm{arccot}}x} \right) = \\ - \frac{1}{{1 + {x^2}}}{\rm{d}}x \end{array}$$

 

函数的和、差、积、商的微分法则

函数的和、差、积、商的求导法则	函数的和、差、积、商的微分法则
\[\left( {u \pm v} \right)' = u' \pm v'\]	\[{\text{d}}\left( {u \pm v} \right) = {\text{d}}u \pm {\text{d}}v\]
\[\left( {Cu} \right)' = Cu'\]	\[{\text{d}}\left( {Cu} \right) = C{\text{d}}u\]
\[\left( {uv} \right)' = u'v + uv'\]	\[{\text{d}}\left( {uv} \right) = v{\text{d}}u + u{\text{d}}v\]
$$\begin{array}{l} \left( {\frac{u}{v}} \right)' = \\ \frac{{u'v - uv'}}{{{v^2}}}(v \ne 0) \end{array}$$	$$\begin{array}{l} {\rm{d}}\left( {\frac{u}{v}} \right) = \\ \frac{{v{\rm{d}}u - {\rm{ud}}v}}{{{v^2}}}(v \ne 0) \end{array}$$

复合函数的微分法则

设$y=f(u)$及$u=g(x)$都可导，则复合函数$y=f[g(x)]$的微分为

\[{\text{d}}y = y{'_x}{\text{d}}x = f'\left( u \right)g'(x){\text{d}}x\]

由于\(g'\left( x \right){\text{d}}x = {\text{d}}u\)

\[{\text{d}}y = y{'_x}{\text{d}}x = f'\left( u \right){\text{d}}u\]

上式表明，无论$u$是中间变量还是自变量，微分形式${\text{d}}y = f'\left( u \right){\text{d}}u$保持不变.这一性质称为微分形式不变性.

 

例 已知微分，求原函数.

$${\text{d}}f\left( x \right) = x{\text{d}}x.$$

解:${\text{d}}\left( {{x^2}} \right) = 2x{\text{d}}x$，所以

$$x{\text{d}}x = \frac{1}{2}{\text{d}}\left( {{x^2}} \right) = {\text{d}}\left( {\frac{1}{2}{x^2}} \right)$$

一般地，有

${\text{d}}\left( {\frac{1}{2}{x^2} + c} \right) = x{\text{d}}x$($c$为任意常数).

 

$${\text{d}}f\left( x \right) = \cos \omega t{\text{d}}t.$$

解:${\text{d}}\left( {\sin \omega t} \right) = \omega \cos \omega t{\text{d}}t$，所以

$$\cos \omega t{\text{d}}t = \frac{1}{\omega }{\text{d}}\left( {\sin \omega t} \right) = {\text{d}}\left( {\frac{1}{\omega }\sin \omega t} \right)$$

一般地，有

${\text{d}}\left( {\frac{1}{\omega }\sin \omega t + c} \right) = \cos \omega t{\text{d}}t$($c$为任意常数).