微分的定义
设函数f(x)f(x)f(x)在x0x_0x0的某个邻域内有定义,若函数增量Δy=f(x0+Δx)−f(x0)\Delta y=f(x_0+\Delta x)-f(x_0)Δy=f(x0+Δx)−f(x0)可表示为Δy=aΔx+o(Δx)\Delta y=a\Delta x+o(\Delta x)Δy=aΔx+o(Δx),其中aaa为不依赖于Δx\Delta xΔx的常数,则称y=f(x)y=f(x)y=f(x)在x0x_0x0处可微,其中aΔxa\Delta xaΔx叫做函数y=f(x)y=f(x)y=f(x)在点x0x_0x0相应于自变量Δx\Delta xΔx的微分,记作dydydy,即dy=aΔxdy=a\Delta xdy=aΔx.
导数和微分的关系
- a=f′(x),Δx=dx,dy=f′(x)dxa=f'(x),\Delta x=dx,dy=f'(x)dxa=f′(x),Δx=dx,dy=f′(x)dx
- f′(x)=dydxf'(x)=\dfrac{dy}{dx}f′(x)=dxdy
- 可导是可微的充要条件
可导是可微的充要条件的证明
充分性
\qquad设f(x)f(x)f(x)在x0x_0x0处可导,则下面极限存在:
limΔx→0f(x0+Δx)−f(x0)Δx=f′(x0)\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x_0+\Delta x)-f(x_0)}{\Delta x}=f'(x_0)Δx→0limΔxf(x0+Δx)−f(x0)=f′(x0)
\qquad即
limΔx→0f(x0+Δx)−f(x0)−f′(x0)ΔxΔx=0\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x_0+\Delta x)-f(x_0)-f'(x_0)\Delta x}{\Delta x}=0Δx→0limΔxf(x0+Δx)−f(x0)−f′(x0)Δx=0
\qquad所以,当Δx→0\Delta x\rightarrow0Δx→0时,
Δf(x0)=f′(x0)Δx+o(Δx)\Delta f(x_0)=f'(x_0)\Delta x+o(\Delta x)Δf(x0)=f′(x0)Δx+o(Δx)
\qquad由此可得f(x)f(x)f(x)在x0x_0x0处可微,并且df(x0)=f′(x0)dxdf(x_0)=f'(x_0)dxdf(x0)=f′(x0)dx.
必要性
\qquad设f(x)f(x)f(x)在x0x_0x0处可微,则∃a∈R\exist a\in R∃a∈R,使得当Δx→0\Delta x\rightarrow0Δx→0时,
Δf(x0)=aΔx+o(Δx)\Delta f(x_0)=a\Delta x+o(\Delta x)Δf(x0)=aΔx+o(Δx)
limΔx→0ΔyΔx=limΔx→0f(x0+Δx)−f(x)Δx=limΔx→0aΔx+o(Δx)Δx=a\lim\limits_{\Delta x\rightarrow0}\dfrac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x_0+\Delta x)-f(x)}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\dfrac{a\Delta x+o(\Delta x)}{\Delta x}=aΔx→0limΔxΔy=Δx→0limΔxf(x0+Δx)−f(x)=Δx→0limΔxaΔx+o(Δx)=a
\qquad即f(x)f(x)f(x)在x0x_0x0处可导,并且f′(x0)=af'(x_0)=af′(x0)=a
例1
设函数f(x)=x3−xf(x)=x^3-xf(x)=x3−x,当x=2,Δx=0.01x=2,\Delta x=0.01x=2,Δx=0.01时,函数yyy的微分dydydy是‾\underline{\qquad}.
解:
f′(x)=(3x2−1)∣x=2=11\qquad f'(x)=(3x^2-1)|_{x=2}=11f′(x)=(3x2−1)∣x=2=11
dy=f′(x)Δx=11×0.01=0.11\qquad dy=f'(x)\Delta x=11\times 0.01=0.11dy=f′(x)Δx=11×0.01=0.11
例2
设函数y=f(x)y=f(x)y=f(x)在x0x_0x0处可微,自变量在x0x_0x0处有改变量Δx=0.2\Delta x=0.2Δx=0.2,相应的函数该变量Δy\Delta yΔy的线性主部等于0.80.80.8,则f′(x0)=‾f'(x_0)=\underline{\qquad}f′(x0)=.
解:
Δy=aΔx+o(Δx)\qquad \Delta y=a\Delta x+o(\Delta x)Δy=aΔx+o(Δx)
\qquad线性主部为dy=aΔx=0.8dy=a\Delta x=0.8dy=aΔx=0.8
\qquad即f′(x0)⋅0.2=0.8f'(x_0)\cdot 0.2=0.8f′(x0)⋅0.2=0.8,得f′(x0)=4f'(x_0)=4f′(x0)=4
总结
利用可导是可微的充要条件,利用导数和微分的关系来解题。