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2.5 函数的微分

§2.5\S 2.5 函数的微分#

引例

S(x)=x2S(x)=x^2

S(x+Δx)=(x+Δx)2=x2+2xΔx+(Δx)2S(x+\Delta x)=(x+\Delta x)^2=x^2+2x\cdot \Delta x+(\Delta x)^2

\begin{align*} \Delta S=S(x+\Delta x)-S(x)&=\underset{线性(函数)部分}{2x\cdot \Delta x}+\underset{高阶无穷小量(\Delta x\to 0)}{(\Delta x)^2}\ &\sim 2x\cdot \Delta x \to 0 \end{align*}

$y=x^3$

\begin{align*} \Delta y=y(x+\Delta x)-y(x)&=(x+\Delta x)^3-x^3\ &=\underset{线性部分}{3x^2\cdot \Delta x}+\underset{o(\Delta x)(\Delta x\to 0)}{3x(\Delta x)^2+(\Delta x)^3} \end{align*}

一、函数的微分#

  • 定义:y=f(x)y=f(x)U(x)U(x) 有定义,若 Δy=f(x+Δx)f(x)=AΔx+o(Δx)\Delta y=f(x+\Delta x)-f(x)=A\cdot \Delta x+o(\Delta x)AAΔx\Delta x 无关),则 y=f(x)y=f(x)xx 处可微,称 AΔxA\cdot \Delta xy=f(x)y=f(x) 的微分,记作 dy=AΔx=Adx\mathrm dy=A\cdot \Delta x=A\cdot \mathrm dx

S=x2S=x^2dS=2xΔx=2xdx\mathrm dS=2x\cdot \Delta x=2x\mathrm dx

y=x3y=x^3dy=3x2Δx=3x2dx\mathrm dy=3x^2\cdot \Delta x=3x^2\mathrm dx

  • 定理:函数在一点可微 \Leftrightarrow 函数在一点可导,且 dy=f(x)dx\mathrm dy=f'(x)\mathrm dx

证明:

\RightarrowΔy=AΔx+o(Δx)\Delta y=A\cdot \Delta x+o(\Delta x)

ΔyΔx=A+o(Δx)Δx\dfrac{\Delta y}{\Delta x}=A+\dfrac{o(\Delta x)}{\Delta x}

f(x)=limΔx0ΔyΔx=A+0=Ady=Adx=f(x)dx\displaystyle f'(x)=\lim_{\Delta x\to 0}\dfrac{\Delta y}{\Delta x}=A+0=A \quad \mathrm dy=A\mathrm dx=f'(x)\mathrm dx

\Leftarrowf(x)=limΔx0ΔyΔxΔyΔx=f(x)+αf'(x)=\displaystyle \lim_{\Delta x\to 0}\dfrac{\Delta y}{\Delta x}\quad \dfrac{\Delta y}{\Delta x}=f'(x)+\alpha

\begin{align*} \Delta y&=f’(x)\Delta x+\alpha \Delta x\ &=\underset{线性部分}{f’(x)\Delta x}+\underset{高阶无穷小}{o(\Delta x)} \end{align*}

例题#

  • y=arctanxy=\arctan xdy=11+x2dx\mathrm dy=\dfrac{1}{1+x^2}\mathrm dxdyx=0=1dx\mathrm dy|_{x=0}=1 \mathrm dx
  • y=x2exy=x^2e^xdy=(2xex+x2ex)dx\mathrm dy=(2xe^x+x^2e^x)\mathrm dx

二、微分的运算法则#

与导数运算法则类似

  • u,vu,v 均可微:
    1. d(u±v)=du±dv\mathrm d(u\pm v)=\mathrm du\pm \mathrm dv
    2. d(uv)=vdu+udv\mathrm d(uv)=v\cdot \mathrm du+u\cdot \mathrm dv
    3. d(uv)=vduudvv2\mathrm d\left(\dfrac{u}{v}\right)=\dfrac{v\cdot \mathrm du-u\cdot \mathrm dv}{v^2}
    4. 一阶微分的形式不变性(设 y=f(u),u=ϕ(x)y=f(u),u=\phi(x) 均可导)
    d{f[ϕ(x)]}=ddxf[ϕ(x)]dx=f(u)ϕdx=f(u)du \begin{align*} \mathrm d\{f[\phi(x)]\}&=\dfrac{\mathrm d}{\mathrm dx}f[\phi(x)]\cdot \mathrm dx\\ &=f'(u)\phi'\mathrm dx\\ &=f'(u)\mathrm du \end{align*}

高阶微分#

dy=f(x)dxdny=f(n)xdxn\mathrm dy=f'(x)\mathrm dx \Rightarrow \mathrm d^ny=f^{(n)}x\mathrm dx^n
  • 高阶导数不具有形式不变性

例题#

  1. y=sin(x2ex)y=\sin(x^2\cdot e^x),求 dy\mathrm dy
dy=cos(x2ex)d(x2ex)=cos(x2ex)[d(x2)ex+x2d(ex)]=[cos(x2ex)(2xex+x2ex)]dx\begin{align*} \mathrm dy&=\cos (x^2e^x)\mathrm d(x^2e^x)\\ &=\cos (x^2e^x)[\mathrm d(x^2)\cdot e^x+x^2\mathrm d(e^x)]\\ &=[\cos (x^2e^x)\cdot (2xe^x+x^2e^x)]\mathrm dx \end{align*}
  1. y=ln(1+ex2)y=\ln (1+e^{x^2}),求 dy\mathrm dy
dy=11+ex2ex22xdx=2xex21+ex2dx\begin{align*} \mathrm dy&=\dfrac{1}{1+e^{x^2}}\cdot e^{x^2}\cdot 2x\mathrm dx\\ &=\dfrac{2xe^{x^2}}{1+e^{x^2}}\mathrm dx \end{align*}

dy=11+ex2d(1+ex2)=2xex21+ex2dx\begin{align*} \mathrm dy&=\dfrac{1}{1+e^{x^2}}\mathrm d(1+e^{x^2})\\ &=\dfrac{2xe^{x^2}}{1+e^{x^2}}\mathrm dx \end{align*}
  1. ey+xye=0e^y+xy-e=0,求 dy\mathrm dy

求导法

eyy+y+xy0=0y=yx+eydy=yx+eydxe^y\cdot y'+y+xy'-0=0 \Rightarrow y'=-\dfrac{y}{x+e^y}\\ \mathrm dy=-\dfrac{y}{x+e^y}\mathrm dx

微分法:两边微分得

eydy+d(xy)ydx+xdy0=0dy=yx+eydx\begin{align*} e^y\mathrm dy+\underset{y\mathrm dx+x\mathrm dy}{\mathrm d(xy)}-0&=0\\ \mathrm dy&=-\dfrac{y}{x+e^y}\mathrm dx \end{align*}
  1. d(tanx)d(sinx)=1cos2xdxcosxdx=1cos3x\dfrac{\mathrm d(\tan x)}{\mathrm d(\sin x)}=\dfrac{\frac{1}{\cos^2x}\mathrm dx}{\cos x\mathrm dx}=\dfrac{1}{\cos ^3x}

三、微分的几何意义#

Δy=f(x0+Δx)f(x)f(x0)f(x0)Δxf(x0)(xx0)=dyx=x0\Delta y=\underset{f(x)}{f(x_0+\Delta x)}-f(x_0)\approx f'(x_0)\cdot \Delta x\\ f'(x_0)(x-x_0)=\mathrm dy|_{x=x_0}
  • y=f(x)y=f(x)x0x_0 处切线方程 yf(x0)=f(x0)(xx0)y-f(x_0)=f'(x_0)(x-x_0)

以直代曲要求

  1. xx00x-x_0\to 0
  2. f(x0),f(x0)f(x_0),f'(x_0) 可计算且相对容易

例题#

  1. sin31\sin 31^\circ 的近似值

解:y=sinxxRy=\sin x \quad x\in R

f(x)f(x0)+f(x0)(xx0)=sinx0+cosx0(xx0)f(x)\approx f(x_0)+f'(x_0)(x-x_0)=\sin x_0+\cos x_0(x-x_0)

sin31sin30+cos30π1800.5+1.7322π180\sin 31^\circ \approx \sin 30^\circ+\cos 30^\circ \cdot \dfrac{\pi}{180}\approx 0.5+\dfrac{1.732}{2}\cdot \dfrac{\pi}{180}

  1. (34)15(34)^{\frac{1}{5}} 的近似值

解:y=x15y=x^{\frac{1}{5}}

34153215+15×3245(3432)=2+15×116×2=2+140=2.025\begin{align*} 34^{\frac{1}{5}}&\approx 32^{\frac{1}{5}}+\dfrac{1}{5}\times 32^{-\frac{4}{5}}(34-32)\\ &=2+\dfrac{1}{5}\times \dfrac{1}{16}\times 2\\ &=2+\dfrac{1}{40}=2.025 \end{align*}
2.5 函数的微分
https://gitee.com/jason_ren/advanced-math-note
作者
Jason Ren
发布于
2025-12-03
许可协议
CC BY-SA 4.0

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