Per aspera ad astra

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微积分笔记

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极限和连续性

何为极限?

假设对于函数 ff 有:

limxcf(x)=L\displaystyle \lim _{x\to c} f(x)=L

即:只要 xx 无限接近于 cc, 则 f(x)f(x) 必然无限接近于 LL

εδ\varepsilon -\delta 语言来描述就是:

ε>0, δ>0, s.t. 0<xc<δf(x)L<ε\displaystyle \forall \varepsilon >0,\ \exists \delta >0,\ s.t.\ 0<|x−c|< \delta \Longrightarrow |f(x)-L|< \varepsilon

说白了就是:无论给定任何一个数字 ε(ε>0)\varepsilon (\varepsilon >0),总能找到一个数 δ(δ>0)\delta ( \delta >0)。使当 xxccδ\delta 范围内时,f(x)f(x) 在极限 LLε\varepsilon 范围内。


例:已知 f(x)={2xx5xx=5f( x) =\begin{cases} 2x & x\neq 5\\ x & x=5 \end{cases}, 证明 limx5f(x)=10\displaystyle \lim _{x\rightarrow 5} f( x) =10

根据定义,给定任意 ε(ε>0)\varepsilon (\varepsilon >0),有 δ(δ>0)\delta ( \delta >0)。因此,我们本质上是要找到一个 δ=function of ε\delta =function\ of\ \varepsilon 的函数。

Proof.\mathnormal{Proof.}

x5<δ2x10<ε2x10<2δ2δ=εδ=ε22x10<εε>0, δ>0, s.t. x5<δ2x10<ε Q.E.D.\begin{aligned} & |x-5| < \delta \Longrightarrow |2x-10|< \varepsilon \\ & |2x-10| < 2\delta \\ & 2\delta =\varepsilon \Rightarrow \delta =\frac{\varepsilon }{2}\\ & |2x-10| < \varepsilon \\ & \forall \varepsilon >0,\ \exists \delta >0 ,\ s.t.\ |x-5|< \delta \Longrightarrow |2x-10|< \varepsilon \ \quad Q.E.D. \end{aligned}

夹逼定理

II 为包含某点 cc 的区间,f,g,hf, g, h 为定义在 II 上的函数。若对于所有属于 II 而不等于 ccxx,有:

则,limxcf(x)=L\displaystyle \lim _{x\rightarrow c} f( x) =L

g(x)g(x)h(x)h(x) 分别被称为 f(x)f(x) 的下界和上界。

Proof: limθ0sinθθ=1\displaystyle \lim _{\theta \rightarrow 0}\frac{\sin \theta }{\theta } =1

1

Proof: limθ01cosθθ=0\displaystyle \lim _{\theta \rightarrow 0}\frac{1-\cos \theta }{\theta } =0

Proof.\mathnormal{Proof.}

limθ01cosθθ=limθ0(1cosθ)(1+cosθ)θ(1+cosθ)=limθ0sin2θθ(1+cosθ)=limθ0sinθθlimθ0sinθ1+cosθ=10=0Q.E.D.\begin{aligned} \lim _{\theta \rightarrow 0}\frac{1-\cos \theta }{\theta } & =\lim _{\theta \rightarrow 0}\frac{( 1-\cos \theta )( 1+\cos \theta )}{\theta ( 1+\cos \theta )}\\ & =\lim _{\theta \rightarrow 0}\frac{\sin^{2} \theta }{\theta ( 1+\cos \theta )}\\ & =\lim _{\theta \rightarrow 0}\frac{\sin \theta }{\theta } \cdot \lim _{\theta \rightarrow 0}\frac{\sin \theta }{1+\cos \theta }\\ & =1\cdot 0\\ & =0 & Q.E.D. \end{aligned}

连续性的定义

函数在某一点处连续:ff is continuous at x=climxcf(x)=f(c)x=c\Longleftrightarrow \displaystyle \lim _{x\rightarrow c} f( x) =f( c)

函数在开区间连续:ff is continuous over (a, b)f( a,\ b) \Longleftrightarrow f is continuous over every point in the interval

函数在闭区间连续:ff is continuous over [a, b]f[ a,\ b] \Longleftrightarrow f is continuous over (a, b)( a,\ b) and limxa+f(x)=f(a)\displaystyle \lim _{x\rightarrow a^{+}} f( x) =f( a), limxbf(x)=f(b)\displaystyle \lim _{x\rightarrow b^{-}} f( x) =f( b)

Intermediate Value Theorem

Suppose ff is a continuous function at every point of the interval [a, b][ a,\ b]

怎么会有这么简单的定理…

导数

导数的两种定义形式

f(x)=limh0f(x+h)f(x)h\displaystyle f^{\prime }( x) =\lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h}

f(c)=limxcf(x)f(c)xc\displaystyle f^{\prime }( c) =\lim _{x\rightarrow c}\frac{f( x) -f( c)}{x-c}

可微性

不可微的三种情况:

  1. not continuous
  2. vertical tangent
  3. “sharp turn”

Proof: Differentiability implies continuity

Proof.\mathnormal{Proof.}

Assume: ff differentiability at x=cx=c

f differentiability at x=cf(c)=limxcf(x)f(c)xc\begin{array}{l} \because f\ differentiability\ at\ x=c\\ \therefore \displaystyle f^{\prime }( c) = \lim _{x\rightarrow c}\frac{f( x) -f( c)}{x-c} \end{array}

limxc[f(x)f(c)]=limxc(xc)f(x)f(c)xc=limxc(xc)limxcf(x)f(c)xc=0f(c)=0limxc[f(x)f(c)]=0limxcf(x)limxcf(c)=0limxcf(x)f(c)=0limxcf(x)=f(c)Q.E.D.\begin{aligned} \lim _{x\rightarrow c}[ f( x) -f( c)] & =\lim _{x\rightarrow c}( x-c) \cdot \frac{f( x) -f( c)}{x-c}\\ & =\lim _{x\rightarrow c}( x-c) \cdot \lim _{x\rightarrow c}\frac{f( x) -f( c)}{x-c}\\ & =0\cdot f^{\prime }( c)\\ & =0\\ & \\ \lim _{x\rightarrow c}[ f( x) -f( c)] & =0\\ \lim _{x\rightarrow c} f( x) -\lim _{x\rightarrow c} f( c) & =0\\ \lim _{x\rightarrow c} f( x) -f( c) & =0\\ \lim _{x\rightarrow c} f( x) & =f( c) & Q.E.D. \end{aligned}

Justifying the power rule

Proof: ddx(xn)=nxn1\displaystyle \frac{d}{dx}\left( x^{n}\right) =nx^{n-1}

Proof.\mathnormal{Proof.}

ddx(xn)=limΔx0(x+Δx)nxnΔx\displaystyle \frac{d}{dx}\left( x^{n}\right) =\lim _{\Delta x\rightarrow 0}\frac{( x+\Delta x)^{n} -x^{n}}{\Delta x}

According to Binomial theorem:

limΔx0(x+Δx)nxnΔx=limΔx0xn+(n1)xn1Δx+(n2)xn2Δx2+...+(nn)x0ΔxnxnΔx=limΔx0(n1)xn1+(n2)xn2Δx+...+(nn)Δxn1=limΔx0(n1)xn1=limΔx0n!1!(n1)!xn1=limΔx0nxn1Q.E.D.\begin{aligned} \displaystyle \lim _{\Delta x\rightarrow 0}\frac{( x+\Delta x)^{n} -x^{n}}{\Delta x} & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\cancel{x^{n}} +\binom{n}{1} x^{n-1} \Delta x+\binom{n}{2} x^{n-2} \Delta x^{2} +...+\binom{n}{n} x^{0} \Delta x^{n}\cancel{-x^{n}}}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\binom{n}{1} x^{n-1} +\cancel{\binom{n}{2} x^{n-2} \Delta x} +...+\cancel{\binom{n}{n} \Delta x^{n-1}}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\binom{n}{1} x^{n-1}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{n!}{\cancel{1!}( n-1) !} x^{n-1}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0} nx^{n-1} & Q.E.D. \end{aligned}

Proof: ddx(x)=12x12\displaystyle \frac{d}{dx}\left(\sqrt{x}\right) =\frac{1}{2} x^{-\frac{1}{2}}

Proof.\mathnormal{Proof.}

ddx(x)=limΔx0x+ΔxxΔx=limΔx0(x+Δxx)(x+Δx+x)Δx(x+Δx+x)=limΔx01x+Δx+x=limΔx012x=limΔx012x12Q.E.D.\begin{aligned} \frac{d}{dx}\left(\sqrt{x}\right) & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sqrt{x+\Delta x} -\sqrt{x}}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\left(\sqrt{x+\Delta x} -\sqrt{x}\right)\left(\sqrt{x+\Delta x} +\sqrt{x}\right)}{\Delta x\left(\sqrt{x+\Delta x} +\sqrt{x}\right)}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{\sqrt{x+\Delta x} +\sqrt{x}}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{2\sqrt{x}}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{2} x^{-\frac{1}{2}} & Q.E.D. \end{aligned}

Justifying the basic derivative rules

Proof: Constant rule (ddxk=0\displaystyle \frac{d}{dx} k=0)

Proof.\mathnormal{Proof.}

k is constanty does not change as x changesf(x+h)f(x)=0ddxk=limh0f(x+h)f(x)h=limh00h=0\begin{array}{l} \because k\ is\ constant\\ \therefore y\ does\ not\ change\ as\ x\ changes\\ \therefore f( x+h) -f( x) =0\\ \therefore \displaystyle \frac{d}{dx} k= \lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h} =\lim _{h\rightarrow 0}\frac{0}{h} =0 \end{array}

Proof: Constant multiple and sum/difference rules

Constant multiple rule: ddx[kf(x)]=kddxf(x)\displaystyle \dfrac{d}{dx}[k\cdot f(x)]=k\cdot\dfrac{d}{dx}f(x)

Sum rule: ddx[f(x)+g(x)]=ddxf(x)+ddxg(x)\displaystyle \dfrac{d}{dx}[f(x)+g(x)]=\dfrac{d}{dx}f(x)+\dfrac{d}{dx}g(x)

Difference rule: ddx[f(x)g(x)]=ddxf(x)ddxg(x)\displaystyle \dfrac{d}{dx}[f(x)-g(x)]=\dfrac{d}{dx}f(x)-\dfrac{d}{dx}g(x)

Proof.\mathnormal{Proof.}

1. f(x)=kg(x)f(x)=kg(x)\displaystyle 1.\ f( x) =kg( x) \Longrightarrow f^{\prime }( x) =kg^{\prime }( x)

f(x)=limh0f(x+h)f(x)h=limh0kg(x+h)kg(x)h=limh0k(g(x+h)g(x)h)=klimh0g(x+h)g(x)h=kg(x)Q.E.D.\begin{aligned} f^{\prime }( x) & =\displaystyle \lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h}\\ & =\displaystyle \lim _{h\rightarrow 0}\frac{kg( x+h) -kg( x)}{h}\\ & =\displaystyle \lim _{h\rightarrow 0} k\left(\frac{g( x+h) -g( x)}{h}\right)\\ & =k\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h}\\ & =kg^{\prime }( x) & Q.E.D. \end{aligned}

2. f(x)=g(x)±j(x)f(x)=g(x)±j(x)\displaystyle 2.\ f( x) =g( x) \pm j( x) \Longrightarrow f^{\prime }( x) =g^{\prime }( x) \pm j^{\prime }( x)

f(x)=limh0g(x+h)±j(x+h)(g(x)±j(x))h=limh0(g(x+h)g(x)h±j(x+h)j(x)h)=limh0g(x+h)g(x)h±limh0j(x+h)j(x)h=g(x)±j(x)Q.E.D.\begin{aligned} f^{\prime }( x) & =\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) \pm j( x+h) -( g( x) \pm j( x))}{h}\\ & =\displaystyle \lim _{h\rightarrow 0}\left(\frac{g( x+h) -g( x)}{h} \pm \frac{j( x+h) -j( x)}{h}\right)\\ & =\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h} \pm \lim _{h\rightarrow 0}\frac{j( x+h) -j( x)}{h}\\ & =g^{\prime }( x) \pm j^{\prime }( x) & Q.E.D. \end{aligned}

Proof: The derivatives of sin(x) and cos(x)

Known limx0sinxx=1\displaystyle \lim _{x\rightarrow 0}\frac{\sin x}{x} =1 and limx01cosxx=0\displaystyle \lim _{x\rightarrow 0}\frac{1-\cos x}{x} =0

Proof.\mathnormal{Proof.}

1. ddx[sinx]=cosx\displaystyle 1.\ \frac{d}{dx}[\sin x] =\cos x

ddx[sinx]=limΔx0sin(x+Δx)sin(x)Δx=limΔx0cosxsinΔx+sinxcosΔxsinxΔx=limΔx0(cosxsinΔxΔx+sinxcosΔxsinxΔx)=limΔx0cosx(sinΔxΔx)+limΔx0sinx(cosΔx1)Δx=cosxlimΔx0sinΔxΔxsinxlimΔx01cosΔxΔx=cosx1sinx0=cosxQ.E.D.\begin{aligned} \frac{d}{dx}[\sin x] & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin( x+\Delta x) -\sin( x)}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\cos x\sin \Delta x+\sin x\cos \Delta x-\sin x}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\left(\frac{\cos x\sin \Delta x}{\Delta x} +\frac{\sin x\cos \Delta x-\sin x}{\Delta x}\right)\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\cos x\left(\frac{\sin \Delta x}{\Delta x}\right) +\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin x(\cos \Delta x-1)}{\Delta x}\\ & =\cos x\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\sin \Delta x}{\Delta x} -\sin x\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1-\cos \Delta x}{\Delta x}\\ & =\cos x\cdot 1-\sin x\cdot 0\\ & =\cos x & Q.E.D. \end{aligned}

2. ddx[cosx]=sinx\displaystyle 2.\ \frac{d}{dx}[\cos x] =-\sin x

1

Proof: The derivative of exe^{x} is exe^{x}

Know the limit definition of e\mathbb{e} is e=limn(1+1n)n=limn0(1+n)1ne=\displaystyle \lim _{n\rightarrow \infty }\left( 1+\frac{1}{n}\right)^{n} =\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{n}}

Proof.\mathnormal{Proof.}

ddx(ex)=limΔx0ex+ΔxexΔx=exlimΔx0eΔx1Δx\begin{aligned} \frac{d}{dx}\left( e^{x}\right) & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{e^{x+\Delta x} -e^{x}}{\Delta x}\\ & =e^{x}\displaystyle \lim _{\Delta x\rightarrow 0}\frac{e^{\Delta x} -1}{\Delta x} \end{aligned}

Let n=eΔx1, we can get n+1=eΔx, such that Δx=ln(n+1) and as Δx0=n0\displaystyle Let\ n=e^{\Delta x} -1,\ we\ can\ get\ n+1=e^{\Delta x} ,\ such\ that\ \Delta x=\ln( n+1) \ and\ as\ \Delta x\rightarrow 0=n\rightarrow 0

We can rewrite to:We\ can\ rewrite\ to:

ddx(ex)=exlimn0nln(n+1)=exlimn01nn1nln(n+1)=exlimn01ln[(1+n)1n]=ex1ln[limn0(1+n)1n]=exQ.E.D.\begin{aligned} \frac{d}{dx}\left( e^{x}\right) & =e^{x}\displaystyle \lim _{n\rightarrow 0}\frac{n}{\ln( n+1)}\\ & =e^{x}\displaystyle \lim _{n\rightarrow 0}\frac{\frac{1}{n} n}{\frac{1}{n}\ln( n+1)}\\ & =e^{x}\displaystyle \lim _{n\rightarrow 0}\frac{1}{\ln\left[( 1+n)^{\frac{1}{n}}\right]}\\ & =e^{x}\frac{1}{\ln\left[\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{n}}\right]}\\ & =e^{x} & Q.E.D. \end{aligned}

Proof: The derivative of ln(x)\ln( x) is 1x\frac{1}{x}

Method 1 (Directly from the definition of the derivative as a limit)

Proof.\mathnormal{Proof.}

ddx(lnx)=limΔx0ln(x+Δx)ln(x)Δx=limΔx0ln(x+Δxx)Δx=limΔx0ln(1+Δxx)Δx=limΔx01Δxln(1+Δxx)=limΔx0ln[(1+Δxx)1Δx]\begin{aligned} \frac{d}{dx}(\ln x) & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln( x+\Delta x) -\ln( x)}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln\left(\frac{x+\Delta x}{x}\right)}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\ln\left( 1+\frac{\Delta x}{x}\right)}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{1}{\Delta x}\ln\left( 1+\frac{\Delta x}{x}\right)\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\ln\left[\left( 1+\frac{\Delta x}{x}\right)^{\frac{1}{\Delta x}}\right] \end{aligned}

Let n=Δxx, Δx=nx, 1Δx=1n1x and as Δx0=n0\displaystyle Let\ n=\frac{\Delta x}{x} ,\ \Delta x=nx,\ \frac{1}{\Delta x} =\frac{1}{n} \cdot \frac{1}{x} \ and\ as\ \Delta x\rightarrow 0=n\rightarrow 0

We can rewrite to:We\ can\ rewrite\ to:

limΔx0ln[(1+Δxx)1Δx]=1xlimn0ln[(1+n)1n]=1xln[limn0(1+n)1n]=1xQ.E.D.\begin{aligned} \displaystyle \lim _{\Delta x\rightarrow 0}\ln\left[\left( 1+\frac{\Delta x}{x}\right)^{\frac{1}{\Delta x}}\right] & =\frac{1}{x}\displaystyle \lim _{n\rightarrow 0}\ln\left[( 1+n)^{\frac{1}{n}}\right]\\ & =\frac{1}{x}\ln\left[\displaystyle \lim _{n\rightarrow 0}( 1+n)^{\frac{1}{n}}\right]\\ & =\frac{1}{x} & Q.E.D. \end{aligned}

Method 2 (Using the fact that ddx(ex)=ex\displaystyle \frac{d}{dx}\left( e^{x}\right) =e^{x} and applying implicit differentiation)

Proof.\mathnormal{Proof.}

Known ddx(ex)=ex\displaystyle Known\ \frac{d}{dx}\left( e^{x}\right) =e^{x}

Let y=ln(x), we can get:\displaystyle Let\ y=\ln( x) ,\ we\ can\ get:

ddx(ey)=ddx(x)eydydx=1dydx=1ey=1elnx=1xQ.E.D.\begin{aligned} \frac{d}{dx}\left( e^{y}\right) & =\frac{d}{dx}( x)\\ e^{y} \cdot \frac{dy}{dx} & =1\\ \frac{dy}{dx} & =\frac{1}{e^{y}}\\ & =\frac{1}{e^{\ln x}}\\ & =\frac{1}{x} & Q.E.D. \end{aligned}

Proof: The product rule

Proof.\mathnormal{Proof.}

ddx[f(x)g(x)]=limh0f(x+h)g(x+h)f(x+h)g(x)+f(x+h)g(x)f(x)g(x)h=limh0[f(x+h)g(x+h)g(x)h+g(x)f(x+h)f(x)h]=[limh0f(x+h)][limh0g(x+h)g(x)h]+[limh0g(x)][limh0f(x+h)f(x)h]=f(x)g(x)+g(x)f(x)Q.E.D.\begin{aligned} \frac{d}{dx}[ f( x) g( x)] & =\displaystyle \lim _{h\rightarrow 0}\frac{f( x+h) g( x+h) -f( x+h) g( x) +f( x+h) g( x) -f( x) g( x)}{h}\\ & =\displaystyle \lim _{h\rightarrow 0}\left[ f( x+h)\frac{g( x+h) -g( x)}{h} +g( x)\frac{f( x+h) -f( x)}{h}\right]\\ & =\left[\displaystyle \lim _{h\rightarrow 0} f( x+h)\right]\left[\displaystyle \lim _{h\rightarrow 0}\frac{g( x+h) -g( x)}{h}\right] +\left[\displaystyle \lim _{h\rightarrow 0} g( x)\right]\left[\displaystyle \lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h}\right]\\ & =f( x) g^{\prime }( x) +g( x) f^{\prime }( x) & Q.E.D. \end{aligned}

Proof: The derivatives of tan(x)\tan( x)cos(x)\cos( x)sec(x)\sec( x) and csc(x)\csc( x)

Proof.\mathnormal{Proof.}

ddx(tanx)=ddx(sinxcosx)ddx(cotx)=ddx(cosxsinx)=cos2x+sin2xcos2x=(sin2x+cos2x)sin2x=1cos2x=1sin2x=sec2x=csc2xddx(secx)=ddx(1cosx)ddx(cscx)=ddx(1sinx)=0cosx+1sinxcos2x=0sinx1cosxsin2x=sinxcos2x=cosxsin2x=tanxsecx=cotxcscxQ.E.D.\begin{aligned} \frac{d}{dx}(\tan x) & =\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) & \frac{d}{dx}(\cot x) & =\frac{d}{dx}\left(\frac{\cos x}{\sin x}\right)\\ & =\frac{\cos^{2} x+\sin^{2} x}{\cos^{2} x} & & =\frac{-\left(\sin^{2} x+\cos^{2} x\right)}{\sin^{2} x}\\ & =\frac{1}{\cos^{2} x} & & =-\frac{1}{\sin^{2} x}\\ & =\sec^{2} x & & =-\csc^{2} x\\ \frac{d}{dx}(\sec x) & =\frac{d}{dx}\left(\frac{1}{\cos x}\right) & \frac{d}{dx}(\csc x) & =\frac{d}{dx}\left(\frac{1}{\sin x}\right)\\ & =\frac{0\cdot \cos x+1\cdot \sin x}{\cos^{2} x} & & =\frac{0\cdot \sin x-1\cdot \cos x}{\sin^{2} x}\\ & =\frac{\sin x}{\cos^{2} x} & & =-\frac{\cos x}{\sin^{2} x}\\ & =\tan x\cdot \sec x & & =-\cot x\cdot \csc x & Q.E.D. \end{aligned}

Proof: The derivatives of axa^{x} (For any positive base a)

Proof.\mathnormal{Proof.}

Known ddx(ex)=ex\displaystyle Known\ \frac{d}{dx}\left( e^{x}\right) =e^{x}

Let a=elna\displaystyle Let\ a=e^{\ln a}

ddx(ax)=ddx[(elna)x]=ddx[e(lna)x]=e(lna)xlna=axlnaQ.E.D.\begin{aligned} \frac{d}{dx}\left( a^{x}\right) & =\frac{d}{dx}\left[\left( e^{\ln a}\right)^{x}\right]\\ & =\frac{d}{dx}\left[ e^{(\ln a) x}\right]\\ & =e^{(\ln a) x} \cdot \ln a\\ & =a^{x} \cdot \ln a & Q.E.D. \end{aligned}

Proof: The derivatives of logax\log_{a} x (For any positive base a1a\neq 1)

Proof.\mathnormal{Proof.}

Known ddx(lnx)=1x\displaystyle Known\ \frac{d}{dx}(\ln x) =\frac{1}{x}

ddx(logax)=ddx(1lnalnx)=1xlnaQ.E.D.\begin{aligned} \frac{d}{dx}(\log_{a} x) & =\frac{d}{dx}\left(\frac{1}{\ln a} \cdot \ln x\right)\\ & =\frac{1}{x\ln a} & Q.E.D. \end{aligned}

Proof: Chain Rule and Quotient Rule

Chain Rule Proof.\mathnormal{Chain\ Rule\ Proof.}

Known: 1. If a function is differentiable, then it is also continuous.2. If function u is continuous at x, then Δu0 as Δx0\begin{aligned} Known:\ & 1.\ If\ a\ function\ is\ differentiable,\ then\ it\ is\ also\ continuous.\\ & 2.\ If\ function\ u\ is\ continuous\ at\ x,\ then\ \Delta u\rightarrow 0\ as\ \Delta x\rightarrow 0 \end{aligned}

For why if function uu is continuous at xx, then Δu0\Delta u\rightarrow 0 as Δx0\Delta x\rightarrow 0:

1

The chain rule tell us: ddx[y(u(x))]=dydx=dydududx\displaystyle The\ chain\ rule\ tell\ us:\ \frac{d}{dx}[ y( u( x))] =\frac{dy}{dx} =\frac{dy}{du} \cdot \frac{du}{dx}

Assuming yy, uu differentiable at xx. We can get:

dydx=limΔx0ΔyΔx=limΔx0ΔyΔuΔuΔx=(limΔx0ΔyΔu)(limΔx0ΔuΔx)=(limΔu0ΔyΔu)(limΔx0ΔuΔx)=dydududxQ.E.D.\begin{aligned} \frac{dy}{dx} & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}\\ & =\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}\\ & =\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta u}\right)\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta u}{\Delta x}\right)\\ & =\left(\displaystyle \lim _{\Delta u\rightarrow 0}\frac{\Delta y}{\Delta u}\right)\left(\displaystyle \lim _{\Delta x\rightarrow 0}\frac{\Delta u}{\Delta x}\right)\\ & =\frac{dy}{du} \cdot \frac{du}{dx} & Q.E.D. \end{aligned}

Quotient Rule Proof.\mathnormal{Quotient\ Rule\ Proof.}

ddx[f(x)g(x)]=ddx[f(x)[g(x)]1]=f[x](g(x))1f[x](g(x))2g(x)=f(x)g(x)f(x)g(x)[g(x)]2=f(x)g(x)f(x)g(x)[g(x)]2Q.E.D.\begin{aligned} \frac{d}{dx}\left[\frac{f( x)}{g( x)}\right] & =\frac{d}{dx}\left[ f( x) \cdot [ g( x)]^{-1}\right]\\ & =f^{\prime }[ x]( g( x))^{-1} -f[ x]( g( x))^{-2} g^{\prime }( x)\\ & =\frac{f^{\prime }( x)}{g( x)} -\frac{f( x) g^{\prime }( x)}{[ g( x)]^{2}}\\ & =\frac{f^{\prime }( x) g( x) -f( x) g^{\prime }( x)}{[ g( x)]^{2}} & Q.E.D. \end{aligned}

Proof: L’Hôpital’s rule

NOTE

This isn’t full proof of L’Hôpital’s rule, just a special case. But it should give some intuition for why it works.

f(a)=0, g(a)=0; f(a) exists, g(a) exists  limxaf(x)g(x)=f(a)g(a)f( a) =0,\ g( a) =0;\ f^{\prime} ( a) \ exists,\ g^{\prime} ( a) \ exists\ \Longleftrightarrow \ \displaystyle \lim _{x\rightarrow a}\frac{f( x)}{g( x)} =\frac{f^{\prime} ( a)}{g^{\prime} ( a)} f(a)g(a)=limxaf(x)f(a)xalimxag(x)g(a)xa=limxaf(x)f(a)g(x)g(a)=limxaf(x)g(x)We know f(a) and g(a) both equal to zeroQ.E.D\begin{aligned} \frac{f^{\prime }( a)}{g^{\prime }( a)} & =\frac{\displaystyle \lim _{x\rightarrow a}\frac{f( x) -f( a)}{x-a}}{\displaystyle \lim _{x\rightarrow a}\frac{g( x) -g( a)}{x-a}} & \\ & =\displaystyle \lim _{x\rightarrow a}\frac{f( x) -f( a)}{g( x) -g( a)} & \\ & =\displaystyle \lim _{x\rightarrow a}\frac{f( x)}{g( x)} & We\ know\ f( a) \ and\ g( a) \ both\ equal\ to\ zero \\ & & Q.E.D \end{aligned}

Mean Value Theorem

If ff is continuous over [a, b][ a,\ b] and every point over (a, b)( a,\ b) is differentiable. Then there exists some c(a, b)c\in ( a,\ b) where ΔyΔx=f(b)f(a)ba=f(c)\displaystyle \frac{\Delta y}{\Delta x} =\frac{f( b) -f( a)}{b-a} =f^{\prime }( c)

Extreme Value Theorem

ff continuous over [a, b] c, d[a, b]:f(c)f(x)f(d)[ a,\ b] \Longrightarrow \exists \ c,\ d\in [ a,\ b] :f( c) \leqslant f( x) \leqslant f( d) for all x[a, b]x\in [ a,\ b]

critical points exists when non endpoint point at x=a {f(a)=0f(a) undefinedx=a\ \begin{cases} f^{\prime }( a) =0\\ f^{\prime }( a) \ undefined \end{cases}

Definite Integral & Riemann Sum

The definite integral of a continuous function ff over the interval [a, b][ a,\ b], denoted by abf(x)dx\displaystyle \int _{a}^{b} f( x) dx, is the limit of a Riemann sum as the number of subdivisions approaches infinity.

abf(x)dx=limni=1nf(xi)Δx\displaystyle \int _{a}^{b} f( x) dx=\lim _{n\rightarrow \infty }\sum _{i=1}^{n} f( x_{i}) \Delta x

Where Δx=ban\displaystyle \Delta x=\frac{b-a}{n} and xi=a+Δxix_{i} =a+\Delta x\cdot i

Definite integrals properties

Sum/Difference:

ab[f(x)±g(x)]dx=abf(x)dx±abg(x)dx\displaystyle \int _{a}^{b}[ f( x) \pm g( x)] dx=\int _{a}^{b} f( x) dx\pm \int _{a}^{b} g( x) dx

Constant multiple:

abkf(x)dx=kabf(x)dx\displaystyle \int _{a}^{b} k\cdot f( x) dx=k\int _{a}^{b} f( x) dx

Reverse interval:

abf(x)dx=baf(x)dx\displaystyle \int _{a}^{b} f( x) dx=-\int _{b}^{a} f( x) dx

Zero-length interval:

aaf(x)dx=0\displaystyle \int _{a}^{a} f( x) dx=0

Adding intervals:

abf(x)dx+bcf(x)dx=acf(x)dx\displaystyle \int _{a}^{b} f( x) dx+\int _{b}^{c} f( x) dx=\int _{a}^{c} f( x) dx

这么简单的东西相信你一定也知道怎么证。那证明就略略略了吧~

First fundamental theorem of calculus

Let ff be a continuous real−valued function defined on [a, b][ a,\ b]. And FF be the function defined, for all xx in [a, b][ a,\ b], by F(x)=axf(t)dt\displaystyle F( x) =\int _{a}^{x} f( t) dt

Then FF is uniformly continuous on [a, b][ a,\ b] and differentiable on the open interval (a, b)( a,\ b), and F(x)=f(x)\displaystyle F^{\prime }( x) =f( x) for all xx in (a, b)( a,\ b) so FF is an antiderivative of ff.

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Second fundamental theorem of calculus / Newton–Leibniz theorem

Let ff be a continuous real−valued function defined on [a, b][ a,\ b] and FF is a continuous function on [a, b][ a,\ b] which is an antiderivative of ff in (a, b)( a,\ b): F(x)=f(x)\displaystyle F^{\prime }( x) =f( x)

If ff is Riemann integrable on [a, b][ a,\ b] then abf(x)dx=F(b)F(a)\displaystyle \int _{a}^{b} f( x) dx=F( b) -F( a)

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Reverse power rule

xndx=xn+1n+1+C, n1\displaystyle \int x^{n} dx=\frac{x^{n+1}}{n+1} +C,\ n\neq -1

Yes that just simple!

Indefinite integration rules

Polynomials

xndx=xn+1n+1+C\displaystyle \int x^{n} dx=\frac{x^{n+1}}{n+1} +C

Radicals

xnmdx=xnm+1nm+1+C\displaystyle \int \sqrt[m]{x^{n}} dx=\frac{x^{\frac{n}{m} +1}}{\frac{n}{m} +1} +C

Trigonometric functions

sin(x)dx=cos(x)+C\displaystyle \int \sin( x) dx=-\cos( x) +C

cos(x)dx=sin(x)+C\displaystyle \int \cos( x) dx=\sin( x) +C

sec2(x)dx=tan(x)+C\displaystyle \int \sec^{2}( x) dx=\tan( x) +C

csc2(x)dx=cot(x)+C\displaystyle \int \csc^{2}( x) dx=-\cot( x) +C

sec(x)tan(x)dx=sec(x)+C\displaystyle \int \sec( x)\tan( x) dx=\sec( x) +C

csc(x)cot(x)dx=csc(x)+C\displaystyle \int \csc( x)\cot( x) dx=-\csc( x) +C

secxdx=lnsecx+tanx+C\displaystyle \int \sec xdx=\ln| \sec x+\tan x| +C(分子分母同乘 secx+tanx\sec x+\tan x

cscxdx=lncscxcotx+C\displaystyle \int \csc xdx=\ln| \csc x-\cot x| +C(分子分母同乘 cscxcotx\csc x-\cot x

下面是另一种方法求这两个不定积分:

secxdx=1cosxdx=cosxcos2xdx=11sin2xcosxdxLet u=sinx=1(1+u)(1u)du=12(11+u+11u)duPartial fractions=12ln1+u1u+C=12ln1+sinx1sinx+C\begin{array}{ l l l } \displaystyle \int \sec xdx & =\displaystyle \int \frac{1}{\cos x} dx & \\ & =\displaystyle \int \frac{\cos x}{\cos^{2} x} dx & \\ & =\displaystyle \int \frac{1}{1-\sin^{2} x}\cos xdx & Let\ u=\sin x\\ & =\displaystyle \int \frac{1}{( 1+u)( 1-u)} du & \\ & =\displaystyle \frac{1}{2}\int \left(\frac{1}{1+u} +\frac{1}{1-u}\right) du & Partial\ fractions\\ & =\displaystyle \frac{1}{2}\ln\left| \frac{1+u}{1-u}\right| +C & \\ & =\displaystyle \frac{1}{2}\ln\left| \frac{1+\sin x}{1-\sin x}\right| +C & \end{array} cscxdx=1sinxdx=sinxsin2xdx=11cos2xsinxdxLet u=cosx=1(1+u)(1u)du=12(11+u+11u)duPartial fractions=12ln1+u1u+C=12ln1+cosx1cosx+C\begin{array}{ l l l } \displaystyle \int \csc xdx & =\displaystyle \int \frac{1}{\sin x} dx & \\ & =\displaystyle \int \frac{\sin x}{\sin^{2} x} dx & \\ & =\displaystyle \int \frac{1}{1-\cos^{2} x}\sin xdx & Let\ u=\cos x\\ & =\displaystyle -\int \frac{1}{( 1+u)( 1-u)} du & \\ & =\displaystyle -\frac{1}{2}\int \left(\frac{1}{1+u} +\frac{1}{1-u}\right) du & Partial\ fractions\\ & =\displaystyle -\frac{1}{2}\ln\left| \frac{1+u}{1-u}\right| +C & \\ & =\displaystyle -\frac{1}{2}\ln\left| \frac{1+\cos x}{1-\cos x}\right| +C & \end{array}

Exponential functions

exdx=ex+C\displaystyle \int e^{x} dx=e^{x} +C

axdx=axln(a)+C\displaystyle \int a^{x} dx=\frac{a^{x}}{\ln( a)} +C

Logarithmic functions

1xdx=lnx+C\displaystyle \int \frac{1}{x} dx=\ln |x|+C

Inverse trigonometric functions

1a2x2dx=arcsin(xa)+C\displaystyle \int \frac{1}{\sqrt{a^{2} -x^{2}}} dx=\arcsin\left(\frac{x}{a}\right) +C

1a2+x2dx=1aarctan(xa)+C\displaystyle \int \frac{1}{a^{2} +x^{2}} dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right) +C

Integration by parts

uvdx=uvdx(uvdx)dx\displaystyle \int uvdx=u\int vdx-\int \left( u^{\prime }\int vdx\right) dx

Integration by reduction formulae

sinnxdx=1nsinn1xcosx+n1nsinn2xdx\displaystyle \int \sin^{n} xdx=-\frac{1}{n}\sin^{n-1} x\cos x+\frac{n-1}{n}\int \sin^{n-2} xdx

cosnxdx=1ncosn1xsinx+n1ncosn2xdx\displaystyle \int \cos^{n} xdx=\frac{1}{n}\cos^{n-1} x\sin x+\frac{n-1}{n}\int \cos^{n-2} xdx

tannxdx=1n1tann1xtann2xdx\displaystyle \int \tan^{n} xdx=\frac{1}{n-1}\tan^{n-1} x-\int \tan^{n-2} xdx

(lnx)ndx=x(lnx)nn(lnx)n1dx\displaystyle \int (\ln x)^{n} dx=x(\ln x)^{n} -n\int (\ln x)^{n-1} dx