第3课:
- 导数的加法法则:(u+v)′=u′+v′(u + v)' = u' + v'(u+v)′=u′+v′
- 导数的数乘法则:(cu)′=cu′(cu)' = cu'(cu)′=cu′
- 可去间断点:A:        g(x)=sin(x)(x),        limx→0sin(x)x=1A: \;\;\;\;g(x) = \frac{sin(x)}{(x)},\;\;\;\;\lim_{x\rightarrow 0}\frac{sin(x)}{x}=1A:g(x)=(x)sin(x),x→0limxsin(x)=1
B:        h(x)=1−cos(x)x,          limx→01−cos(x)x=0B:\;\;\;\; h(x) = \frac{1-cos(x)}{x},\;\;\;\;\;\lim_{x\rightarrow 0}\frac{1-cos(x)}{x} = 0B:h(x)=x1−cos(x),x→0limx1−cos(x)=0 - sin(x)′=cos(x)sin(x)' = cos(x)sin(x)′=cos(x)和(cosx)′=−sin(x)(cosx)'= -sin(x)(cosx)′=−sin(x)的证明:sin(x+Δx)−sin(x)Δx=sinxcosΔx+cosxsinΔx−sinxΔx=sinx(cosΔx−1Δx)+cosx(sinΔxΔx)=cosx−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−cos(x+Δx)−cosxΔx=cosxcosΔx−sinxsinΔx−cosxΔx=cosx(cosΔx−1Δx)−sinx(sinΔxΔx)=−sinx\frac{sin(x+\Delta x) - sin(x)}{\Delta x} = \frac{sinxcos\Delta x + cosxsin\Delta x - sinx}{\Delta x}\\=sinx(\frac{cos \Delta x - 1}{\Delta x}) + cosx(\frac{sin \Delta x}{\Delta x})\\=cosx \\ --------------------------------------\\ \frac{cos(x + \Delta x)- cos x}{\Delta x}=\frac{cosxcos\Delta x - sinx sin \Delta x - cosx}{\Delta x}\\=cosx(\frac{cos \Delta x - 1}{\Delta x})- sinx(\frac{sin \Delta x}{\Delta x})\\=-sinxΔxsin(x+Δx)−sin(x)=ΔxsinxcosΔx+cosxsinΔx−sinx=sinx(ΔxcosΔx−1)+cosx(ΔxsinΔx)=cosx−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−Δxcos(x+Δx)−cosx=ΔxcosxcosΔx−sinxsinΔx−cosx=cosx(ΔxcosΔx−1)−sinx(ΔxsinΔx)=−sinx
1. 使用极限几何证明A、B。
(1)证明 A,如图所示的单位圆:
Δx→Θ,                弓弦弧长=2sinΘ2Θ→1\Delta x \rightarrow \Theta,\;\;\;\;\;\;\;\; \frac{弓弦}{弧长} = \frac{2 sin \Theta}{2 \Theta} \rightarrow 1Δx→Θ,弧长弓弦=2Θ2sinΘ→1
注解:极短的曲线可以看作直线。弧长公式:弧长=n∗π∗r1800弧长 = \frac{n*\pi*r}{180^0}弧长=1800n∗π∗r,所以角度θ\thetaθ对应的弧长为 θ\thetaθ。
(2)证明B,如图所示的单位圆:
1−cosΘΘ→0,\frac{1-cos\Theta}{\Theta}\rightarrow 0,Θ1−cosΘ→0,前提:这个圆还是单位圆。当红色部分,即 1−cosΘ1- cos \Theta1−cosΘ 很小时,即Θ\ThetaΘ 很小的时候,等式便为0。
(3) 证明:dsinθΔΘ=cosΘ\frac{d sin \theta}{ \Delta \Theta} = cos\ThetaΔΘdsinθ=cosΘ
其sinθsin\thetasinθ的变化可以看作是P、QP、QP、Q两点的纵坐标的变化,即看作是Δy\Delta yΔy。
将OTOTOT和OPOPOP同时旋转90090^0900,相当于,OTOTOT变成了PRPRPR,OPOPOP变成了PQPQPQ,则∠QPR=θ\angle QPR = \theta∠QPR=θ。
OR⊥PQ,QR⊥OP,Δy=PR,PQ^(弧长)=PQˉ(直线)≈ΔΘ,ΔyΔΘ=cosΘOR \perp PQ ,QR \perp OP ,\Delta y = PR, \hat{PQ}(弧长) = \bar{PQ}(直线) \approx \Delta \Theta,\frac{\Delta y}{ \Delta \Theta} = cos\ThetaOR⊥PQ,QR⊥OP,Δy=PR,PQ^(弧长)=PQˉ(直线)≈ΔΘ,ΔΘΔy=cosΘ
2. 补充
- 导数的除法法则:(uv)′=(u′v−uv′)v2                (v≠0)(\frac{u}{v})' = \frac{(u'v - uv')}{v^2}\;\;\;\;\;\;\;\;(v \neq 0)(vu)′=v2(u′v−uv′)(v̸=0)
- 导数的乘法法则:(uv)′=(u′v+uv′)(uv)'=(u'v+uv')(uv)′=(u′v+uv′)
Pf:Pf:Pf:
Δ(uv)=u(x+Δx)v(x+Δx)−u(x)v(x)=(u(x+Δx)−u(x))v(x+Δx)+u(x)(v(x+Δx)−v(x))=Δuv(x+Δx)+u(x)ΔvΔ(uv)Δx=ΔuΔxv(x+Δx)+uΔvΔx由于Δx→0所以,d(uv)dx=dudx⋅v+udvdx\Delta (uv) = u(x + \Delta x)v(x+\Delta x) - u(x)v(x)\\=(u(x+\Delta x) - u(x))v(x+\Delta x)+u(x)(v(x+\Delta x)- v(x))\\=\Delta uv(x+\Delta x)+u(x)\Delta v \\ \frac {\Delta (uv)}{\Delta x}= \frac{\Delta u}{\Delta x}v(x + \Delta x) + \frac{u \Delta v}{ \Delta x}\\由于\Delta x \rightarrow 0 所以,\\ \frac{d(uv)}{dx}=\frac{du}{dx} \cdot v+ u \frac{dv}{dx}Δ(uv)=u(x+Δx)v(x+Δx)−u(x)v(x)=(u(x+Δx)−u(x))v(x+Δx)+u(x)(v(x+Δx)−v(x))=Δuv(x+Δx)+u(x)ΔvΔxΔ(uv)=ΔxΔuv(x+Δx)+ΔxuΔv由于Δx→0所以,dxd(uv)=dxdu⋅v+udxdv
Δ(ux)=u+Δxv+Δv−uv=uv+(Δu)v−uv−uΔv(v+Δv)v=(Δu)v−uΔv(v+Δv)vΔ(ux)Δx=ΔuΔxv−uΔvΔx(v+Δv)v由于Δx→0,所以ddx(uv)=dudx⋅v−udvdxv⋅v\Delta (\frac{u}{x})=\frac{u + \Delta x}{v + \Delta v}-\frac{u}{v}\\=\frac{uv+(\Delta u)v- uv - u\Delta v}{(v + \Delta v)v}\\=\frac{(\Delta u)v- u \Delta v}{(v + \Delta v)v}\\\frac{\Delta(\frac{u}{x})}{\Delta x}=\frac{\frac{\Delta u}{\Delta x}v-u\frac{\Delta v}{\Delta x}}{(v + \Delta v )v}\\由于\Delta x \rightarrow 0,所以\\\frac{d}{dx}(\frac{u}{v})=\frac{\frac{du}{dx} \cdot v - u \frac{dv}{dx}}{v \cdot v}Δ(xu)=v+Δvu+Δx−vu=(v+Δv)vuv+(Δu)v−uv−uΔv=(v+Δv)v(Δu)v−uΔvΔxΔ(xu)=(v+Δv)vΔxΔuv−uΔxΔv由于Δx→0,所以dxd(vu)=v⋅vdxdu⋅v−udxdv