求导微分即求函数的变化率变化趋势
一元函数微分
导数概念
$$
f’(x_0)=\underset{\bigtriangleup x \rightarrow 0}{\lim}\dfrac{\bigtriangleup y}{\bigtriangleup x}=\underset{\bigtriangleup x \rightarrow 0}{\lim}\dfrac{f(x+\bigtriangleup x_0)-f(x_0)}{\bigtriangleup x}
$$
导数性质
可导一定连续；连续不一定可导
可导一定可微；可微一定可导
可微一定连续；连续不一定可微
- $\left. \begin{array}{r} \exists f_+'\left( x_0 \right)\\ \exists f_-'\left( x_0 \right)\\ \end{array} \right\} \Rightarrow f\left( x \right) \in \text{C}_{x=x_0}$
对于$f(x)=F(x)(x-a)^\alpha$在$x=a$处导数
- $\alpha>1$，导数存在且为$0$
- $\alpha<1$，导数不存在
- $\alpha=1$，无绝对值则可导，有绝对值$F(a)=0$时也可导
求导
基本求导公式
$\left( x^a \right)’=ax^{a-1}$
$\left( a^x \right)’=a^x\ln a$
$\left( \text e^x \right)’=\text e^x$
$\left( \log_ax \right)’=\dfrac{1}{x\ln a}$
$\left( \ln\text{|}x| \right)’=\dfrac{1}{x}$
$\left( \sin x \right)’=\cos x$
$\left( \cos x \right)’=-\sin x$
$\left( \tan x \right)’=\sec ^2x$
$\left( \sec x \right)’=\sec x\tan x$
$\left( \csc x \right)’=-\csc x\cot x$
$\left( \cot x \right)’=-csc^2x$
$\left( \arcsin x \right)’=\dfrac{1}{\sqrt{1-x^2}}$
$\left( \arccos x \right)’=-\dfrac{1}{\sqrt{1-x^2}}$
$\left( \arctan x \right)’=\dfrac{1}{1+x^2}$
$\left( \text {arccot} x \right)’=-\dfrac{1}{1+x^2}$
$\ln \left( x+\sqrt{x^2+1} \right)’=\dfrac{1}{\sqrt{x^2+1}}$
$\ln \left( x-\sqrt{x^2+1} \right)’=\dfrac{1}{\sqrt{x^2-1}}$
$\ln \left( x+\sqrt{x^2+a} \right) =\dfrac{1}{\sqrt{x^2+a^2}}$
反函数导数
$\dfrac{\text{d}x}{\text{d}y}=\dfrac{1}{f’(x)}$
$\dfrac{\text{d}^2x}{\text{d}y^2}=-\dfrac{y’’}{(y’)^3}$
$\dfrac{\text{d}^3x}{\text{d}y^3}=\dfrac{3y’’^2-y’y’’’}{y’^5}$
参数方程导数
$\dfrac{\text dy}{\text dx}=\dfrac{y’_t}{x’_t}$
$\dfrac{\text d^2y}{\text dx^2}=\dfrac{\text d(\dfrac{\text dy}{\text dx})/\text dt}{\text dx/\text dt}=\dfrac{y’’_tx’_t-y’_tx’’_t}{(x’_t)^3}$
隐函数求导
$F(x,y)=0$，则$\dfrac{dy}{dx}=-\dfrac{F_x’}{F_y’}$
对数求导：$\dfrac{y’}{y}=y (\ln y)’$
幂指求导：
$(u^v)’=u^v(v’\ln u+\dfrac {vu’}u)^2$
$(u^v)’’’=u^v(v’\ln u+\dfrac {vu’}u)^2+v’’’\ln u+\dfrac{2u’v’}{u}+\dfrac{v(uu’’-u’^2)}{u^2}$
导数应用
判断极值
$\underset{x\rightarrow a}{\lim}\dfrac{f\left( x \right) -f\left( a \right)}{\left( x-a \right) ^2}\left\{ \begin{array}{l}>0\ \ \ \ \ \text{极小}\\ <0\ \ \ \ \ \text{极大}\\ \end{array} \right. $判断不可导
$\underset{x\rightarrow 0}{\lim}\dfrac{f(x+h)-f(k-x)}{g(x)}$可能在$x=k$处无定义
$\underset{x\rightarrow 0}{\lim}\dfrac{f(k+\varphi(x))-f(k)}{g(x)} $若$\varphi(x)、g(x)\ge0$，则只能单侧可导
高阶导数
性质 & 求法
- $f(x)=x^k|x|(k>1)$，则使$f^{(n)}(0)$存在的最高阶数为$k$
- $f^{(k)}(0)=k!a_k$ ($a_k$为第$k$项系数)
高阶求导公式(莱布尼茨公式)： $(u+v)^{(n)}=u^{(n)}+v^{(n)}\\ (uv)^{(n)}=\sum ^{n} _{k=0} C_n^ku^{(n-k)}v^{(k)}$
泰勒公式(麦克劳林展开式)：$f(x)=\sum ^∞ _{n=0} \dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n\\ f(x)=\sum ^∞ _{n=0} \dfrac{f^{(n)}(0)}{n!}x^n$
$y=x^nf(x)$求$y^{(n)}(0)$
- 由于$y=x^nf(x)$无穷阶可导，则可将其抽象展开为$y=\sum_{n=0}^{\infty}\dfrac{y^{(n)}(0)}{n!}x^n$
- 又因为$y=x^nf(x)=$展开
- 根据函数展开的唯一性，比较1、2公式中的系数，则$\dfrac{y^{(n)}(0)}{n!}=(找2中展开的对应次项)$
常见高阶导数
$\left( \text{e}^x \right) ^{\left( n \right)}=\text{e}^x$
$\left( \text{e}^{f\left( x \right)} \right) ^{\left( n \right)}=n!\text{e}^{nf\left( x \right)}$
$\left( a^x \right) ^{\left( n \right)}=a^x\left( \ln a \right) ^n$
$\left( \text{e}^{ax+b} \right) ^{\left( n \right)}=a^n\text{e}^{ax+b}$
$\left[ \sin \left( ax+b \right) \right] ^{\left( n \right)}=a^n\sin \left( ax+b+\dfrac{n\pi}{2} \right) $
$\left[ \cos \left( ax+b \right) \right] ^{\left( n \right)}=a^n\cos \left( ax+b+\dfrac{n\pi}{2} \right) $
$(\arctan x)^{(n)}=(n-1)!\cos^ny\cos[ny+\dfrac{(n-1)\pi}2]$
$\left[ \ln \left( ax+b \right) \right] ^{\left( n \right)}=\left( -1 \right) ^{n-1}a^n\dfrac{\left( n-1 \right) !}{\left( ax+b \right) ^n}$
$\left( \dfrac{1}{ax+b} \right) ^{\left( n \right)}=\left( -1 \right) ^na^n\dfrac{n!}{\left( ax+b \right) ^{n+1}}$
$\left[ \left( x+x_0 \right) ^m \right] ^{\left( n \right)}=m\left( m-1 \right) \left( m-2 \right) …\left( m-n+1 \right) \left( x+x_0 \right) ^{m-n}$
$f(x)=\alpha f_1(x)+\beta f_2(x) \Rightarrow f^{\left( n \right)}\left( x \right) =\alpha f_1^{\left( n \right)}\left( x \right) +\beta f_2^{\left( n \right)}\left( x \right) $$\left( uv \right) ^{\left( n \right)}=\sum{\text{C}_{n}^{k}u^{\left( n-k \right)}v^{\left( k \right)}}$
多元函数微分
连续 & 偏导 & 可微
极限：$|f(x,y)-A|<\varepsilon$
连续：$\underset{y\rightarrow y_0}{\underset{x\rightarrow x_0}{\lim}}f(x_0,y_0)=f(x_0,y_0)$
偏导数：$f_x’(x_0,y_0)=\left. \frac{\partial f}{\partial x} \right|\underset{y=y_0}{_{x=x_0}}=\underset{\bigtriangleup x\rightarrow 0}{\lim}\dfrac{f\left( x_0+\bigtriangleup x,y_0 \right) -f\left( x_0,y_0 \right)}{\bigtriangleup x}$
可微：$\bigtriangleup S=y \bigtriangleup x+ x \bigtriangleup y$
判断连续
若有间断点$(0,0)$，可令$\left\{ \begin{array}{l} x=\rho \cos \theta\\ y=\rho \sin \theta\\ \end{array} \right. $
- $f=\dfrac{x^py^q}{x^m+y^n}$ $\dfrac{p}{m}+\dfrac{q}{n}\left\{ \begin{array}{l}>1 \ \ \ \ f\left( 0,0 \right) =0\\ \le 1 \ \ \ \ f\left( 0,0 \right) \text{不存在}\\ \end{array} \right. $
凑$\dfrac{\sin \Box}\Box$
$\sin x<x$
- $\left. \begin{array}{r} x^2+y^2\\ xy\\ \end{array} \right\} \Rightarrow x^2+y^2\ge 2xy$
积分不存在
- 找到两种不同趋近方式，使$\lim f(x,y)$存在但不相等
- 特殊趋向：$y=x$、$y=x^2$
求偏导
求 $f(x_0,y_0)$ 偏导先带后求：$f'_{x_0}(x_0,y_0)=f'(x,y_0)|_{x=x_0}$
偏导数连续
- 定义法求$f_x’(x_0,y_0)$，$f_y’(x_0,y_0)$
- 公式法求$f_x’(x,y)$，$f_y’(x,y)$
- 计算$\left\{ \begin{array}{l} \underset{\begin{array}{c} \bigtriangleup x\rightarrow x_0\\ \bigtriangleup y\rightarrow y_0\\ \end{array}}{\lim}f'_x\left( x,y \right) =f'_x\left( x_0,y_0 \right)\\ \underset{\begin{array}{c} \bigtriangleup x\rightarrow x_0\\ \bigtriangleup y\rightarrow y_0\\ \end{array}}{\lim}f'_y\left( x,y \right) =f'_y\left( x_0,y_0 \right)\\ \end{array} \right. $，若都成立，则$f(x,y)$在$(x_0,y_0)$偏导数连续
判断可微
- 求$f_{x}^{‘}\left( x_0,y_0 \right) $与$f_{y}^{‘}\left( x_0,y_0 \right) $，若有一个不存在则直接不可微
- 作全增量$\bigtriangleup z=f\left( x_0+\bigtriangleup x,y_0+\bigtriangleup y \right) -f\left( x_0,y_0 \right) $
- 表示为线性增量$A\bigtriangleup x+B\bigtriangleup y$，其中$A=f_x’(x_0,y_0)$，$B=f_y’(x_0,y_0)$
- 作极限$\underset{\begin{array}{c} \bigtriangleup x\rightarrow x_0\\ \bigtriangleup y\rightarrow y_0\\ \end{array}}{\lim}\dfrac{\bigtriangleup z-\left( A\bigtriangleup x+B\bigtriangleup y \right)}{\sqrt{\left( \bigtriangleup x \right) ^2+\left( \bigtriangleup y \right) ^2}}\left\{ \begin{array}{l} =0\ z=f\left( x,y \right) \text{在}\left( x_0,y_0 \right) \text{可微}\\ \ne 0\ z=f\left( x,y \right) \text{在}\left( x_0,y_0 \right) \text{不可微}\\ \end{array} \right. $
- $f=\dfrac{x^py^q}{x^m+y^n}\Rightarrow\dfrac{p+q}{m+1}=\left\{ \begin{array}{l}>1\left( 0,0 \right) \ \ \ \ \text{可微}\\ \le 1\left( 0,0 \right) \ \text{不可微}\\ \end{array} \right. $
多元函数求导
链式求导
设$z=f(u,v),u=\varphi(t),v=\phi(t)，则\dfrac{dz}{dt}=\dfrac{\partial z}{\partial u}\dfrac{du}{dt}+\dfrac{\partial z}{\partial v}\dfrac{dv}{dt}$
二阶导
| 原函数 | 一阶导 | 二阶导 |
|---|---|---|
| $f\left(u(x),v(x)\right)$ | $f'=u'f'_1+v'f'_2$ | $f''=u''f_1'+v''f'_2+u'^2f''_{11}+2u'v'f''_{12}+v'^2f''_{12}$ | |
| $f\left(u(x,y),v(x,y)\right)$ | $f''_{xy}=u''_{xy}f_1'+v_{xy}f_2'\\+u_x'u_y'f''_{11}+(u_x'v_y'+u'_yv'_x)f''_{12}+v_x'v_y'f_{22}''$ |
隐函数求导
$\left| \begin{matrix}
\dfrac{\partial F}{\partial u}& \dfrac{\partial F}{\partial v}\
\dfrac{\partial G}{\partial u}& \dfrac{\partial G}{\partial v}\
\end{matrix} \right|=\dfrac{\partial \left( F,G \right)}{\partial \left( u,v \right)}$
雅可比行列式
$\text{d}x\text{d}y=|J|\text{d}u\text{d}v$
$\text{d}\alpha \text{d}\beta \text{d}\gamma =|\dfrac{\partial \left( \alpha ,\beta ,\gamma \right)}{\partial \left( x,y,z \right)}|\text{d}x\text{d}y\text{d}z$
二元方程：$F(x,y,z)=0$，则
$$ \dfrac{\partial z}{\partial x}=-\dfrac{F'_x}{F'_{\boldsymbol{z}}}\ \ \ \ \dfrac{\partial z}{\partial y}=-\dfrac{F'_y}{F'_{\boldsymbol{z}}} $$三元方程组：$\left\{ \begin{array}{l} F\left( x,y,z \right) =0\\ G\left( x,y,z \right) =0\\ \end{array} \right. $，则
$$ \dfrac{\text{d}y}{\text{d}x}=-\dfrac{\dfrac{\partial \left( F,G \right)}{\partial \left( x,z \right)}}{\dfrac{\partial \left( F,G \right)}{\partial \left( y,z \right)}}\ \ \ \ \ \ \ \ \dfrac{\text{d}z}{\text{d}x}=-\dfrac{\dfrac{\partial \left( F,G \right)}{\partial \left( y,x \right)}}{\dfrac{\partial \left( F,G \right)}{\partial \left( y,z \right)}} $$四元方程组：$\left\{ \begin{array}{l} F\left( x,y,u,v \right) =0\\ G\left( x,y,u,v \right) =0\\ \end{array} \right. $，则
$$ \dfrac{\partial u}{\partial x}=-\dfrac{\dfrac{\partial \left( F,G \right)}{\partial \left( x,v \right)}}{\dfrac{\partial \left( F,G \right)}{\partial \left( u,v \right)}}\ \ \ \ \ \ \ \ \dfrac{\partial v}{\partial x}=-\dfrac{\dfrac{\partial \left( F,G \right)}{\partial \left( u,x \right)}}{\dfrac{\partial \left( F,G \right)}{\partial \left( u,v \right)}} \\ \dfrac{\partial u}{\partial y}=-\dfrac{\dfrac{\partial \left( F,G \right)}{\partial \left( y,v \right)}}{\dfrac{\partial \left( F,G \right)}{\partial \left( u,v \right)}}\ \ \ \ \ \ \ \ \dfrac{\partial v}{\partial y}=-\dfrac{\dfrac{\partial \left( F,G \right)}{\partial \left( u,y \right)}}{\dfrac{\partial \left( F,G \right)}{\partial \left( u,v \right)}} $$极值与最值
多元函数泰勒公式
$f\left( x,y \right) =f\left( x_0,\ y_0 \right) \\ +f_{x}^{'}\left( x_0,\ y_0 \right) \left( x-x_0 \right) +f_{y}^{'}\left( x_0,\ y_0 \right) \left( y-y_0 \right) \\ +\dfrac{1}{2}\left( f_{xx}^{''}\left( x_0,\,\,y_0 \right) \left( x-x_0 \right) ^2+2f_{xy}^{''}\left( x_0,\,\,y_0 \right) \left( x-x_0 \right) \left( y-y_0 \right) +f_{yy}^{''}\left( x_0,\,\,y_0 \right) \left( y-y_0 \right) ^2 \right) $无条件极值
必要条件
$\left\{ \begin{array}{l} f_{x}^{'}\left( x_0,y_0 \right) =0\\ f_{y}^{'}\left( x_0,y_0 \right) =0\\ \end{array} \right.$充分条件
$\left\{ \begin{array}{l} f_{xx}^{''}\left( x_0,y_0 \right) =A\\ f_{xy}^{''}\left( x_0,y_0 \right) =B\\ f_{yy}^{''}\left( x_0,y_0 \right) =C\\ \end{array} \right. $，则$\bigtriangleup =B^2-AC=\left\{ \begin{array}{l} <0\ \ \text{极值 }\left\{ \begin{array}{l} A<0,\ \text{极大值}\\ A>0,\ \text{极小值}\\ \end{array} \right.\\ =0\ \ \text{失效，取特殊路径}\\>0\ \ \text{非极值}\\ \end{array} \right.$条件极值
求目标函数$u=f(x,y,z)$在条件$\left\{ \begin{array}{l} \varphi \left( x,y,z \right) =0\\ \psi \left( x,y,z \right) =0\\ \end{array} \right. $下的最值- 构造辅助函数$F(x,y,z,\lambda,\mu)=f(x,y,z)+\lambda \varphi(x,y,z)+\mu \psi(x,y,z)$
- 令$\left\{ \begin{array}{l} \begin{array}{l} F_{x}^{'}\\ \end{array}=f_{x}^{'}+\lambda \varphi _{x}^{'}+\mu \psi _{x}^{'}=0\\ \begin{array}{l} F_{y}^{'}\\ \end{array}=f_{y}^{'}+\lambda \varphi _{y}^{'}+\mu \psi _{y}^{'}=0\\ \begin{array}{l} F_{z}^{'}\\ \end{array}=f_{z}^{'}+\lambda \varphi _{z}^{'}+\mu \psi _{z}^{'}=0\\ \begin{array}{l} F_{\lambda}^{'}\\ \end{array}=\varphi \left( x,y,z \right) =0\\ \begin{array}{l} F_{\mu}^{'}\\ \end{array}=\psi \left( x,y,z \right) =0\\ \end{array} \right.$
- 解上述方程组得备选点$P_i$，并求$u_{max}=max(P_i),u_{min}=min(P_i)$
