MATH1013总结
一些定理与知识点的总结
- 三角函数
- 几个新的三角函数
- 有关反三角函数的计算
- 三角双曲函数
- 和差化积与积化和差公式
- 公式表述
- 公式证明
- 极限
- 某些重要定理(及其可能比较重要的极限)
- Sandwich Theorem
- 定理表述
- 定理的证明
- 定理的应用
- 某些重要极限及其证明
- 写在最前
- e=limx→∞(1+1x)xe= \lim_{x\rightarrow \infty}(1+\frac{1}{x})^{x}e=limx→∞(1+x1)x
- 一些三角函数的极限
- limx→asinx=sina\lim_{x \rightarrow a}sin\ x=sin\ alimx→asin x=sin a
- limx→0sinxx=1\lim_{x \rightarrow 0}\frac{sin\ x}{x}=1limx→0xsin x=1
- limx→0x2sin1x=0\lim_{x \rightarrow 0}x^{2}sin\ \frac{1}{x}=0limx→0x2sin x1=0
- solution to the APPLICATION of important limits
- 导数
- 隐函数求导
- 高中的simple task
- 大学的hard task(雾)
- 新的思考
- 反函数与原函数的导数关系
- 利用导数进行估算
- 与绝对值有关
- local maximum and absolute maximum
- Tricky Question
- Solution to Tricky Question
三角函数
几个新的三角函数
我们在1013的课堂上引入了三个新的三角函数:
cscx=1sinxsecx=1cosxcotx=1tanxcsc\ x\ =\ \frac{1}{sin\ x} \\ sec\ x\ =\ \frac{1}{cos\ x} \\ cot\ x\ =\ \frac{1}{tan\ x} csc x = sin x1sec x = cos x1cot x = tan x1
以及有关反三角函数
arcsinx=sin−1xarccosx=cos−1xarctanx=tan−1xarccscx=csc−1xarcsecx=cos−1xarccotx=cot−1xwhenx∈somesetsarcsin\ x\ =\ sin^{-1}\ x\ \\ arccos\ x\ =\ cos^{-1}\ x\ \\ arctan\ x\ =\ tan^{-1}\ x\ \\ arccsc\ x\ =\ csc^{-1}\ x\ \\ arcsec\ x\ =\ cos^{-1}\ x\ \\ arccot\ x\ =\ cot^{-1}\ x\ \\ when\ x\ \in\ some\ sets arcsin x = sin−1 x arccos x = cos−1 x arctan x = tan−1 x arccsc x = csc−1 x arcsec x = cos−1 x arccot x = cot−1 x when x ∈ some sets
反三角函数与三角函数的反函数其实并不一样,因为根据反函数的定义,只有所谓onetoonefunctionone\ to\ one\ functionone to one function才有反函数,所以三角函数的反函数只能是反三角函数的一部分。通常三角函数的反函数值域都是从原点开始考虑。因为反函数的值域是原函数的定义域,反函数的定义域是原函数的值域。
有关反三角函数的计算
有关反三角函数与三角函数嵌套的计算,绝大部分思路其实都是直接画出一个三角形来计算,比如:
计算:arcsin(cosθ)?计算:arcsin(cos\ \theta)? 计算:arcsin(cos θ)?
我们现在脑补一个直角三角形,主要是图片太麻烦博主懒得搞了。它有一个非直角的角,我们记这个角为θ\thetaθ,于是我们设出以及脑补出这个三角形的一些参数来:
θopposite:xθadjacent:1θhypotenuse:x2+1anotherangle:π2−θ\theta\ opposite:x \\\theta\ adjacent:1 \\\theta\ hypotenuse: \sqrt{x^{2}+1} \\\ another\ angle:\frac{\pi}{2} - \theta θ opposite:xθ adjacent:1θ hypotenuse:x2+1 another angle:2π−θ
于是乎我们分别表示一下:
cosθ=1x2+1soweneedtosolvethefollowingequation:sinθ′=1x2+1soθ′=π2−θcos\ \theta\ =\ \frac{1}{\sqrt{x^{2}+1}} \\so\ we\ need\ to\ solve\ the\ following\ equation: \\ sin \ \theta^{'}\ =\ \frac{1}{\sqrt{x^{2}+1}} \\so\ \theta^{'}\ =\ \frac{\pi}{2} - \theta cos θ = x2+11so we need to solve the following equation:sin θ′ = x2+11so θ′ = 2π−θ
如果是三角函数在外也是同理:
Calculate:sin(arccos1x2+1)?origianlequation=sinθ=xx2+1Calculate:sin(arccos\ \frac{1}{\sqrt{x^{2}+1}})? \\origianl\ equation\ =\ sin\ \theta\ =\ \frac{x}{\sqrt{x^{2}+1}} Calculate:sin(arccos x2+11)?origianl equation = sin θ = x2+1x
三角双曲函数
仅仅是在某次tutorial的习题中遇见了三角双曲函数,但上课也没提过,然后tutorial也就开学的时候去过一次,也不知道之后有没有再出现过,于是当前阶段也只是把公式搬运一遍留个印象而已。
首先我们要知道两个正弦与余弦两个双曲函数:
sinhx=ex−e−x2coshx=ex+e−x2sinh\ x\ =\ \frac{e^{x}-e^{-x}}{2} \\ cosh\ x\ =\ \frac{e^{x}+e^{-x}}{2} sinh x = 2ex−e−xcosh x = 2ex+e−x
然后其他的函数也就呼之欲出了:
tanhx=sinhxcoshxcschx=1sinhxsechx=1coshxcothx=1tanhxtanh\ x\ =\ \frac{sinh\ x}{cosh\ x} \\csch\ x\ =\ \frac{1}{sinh\ x} \\sech\ x\ =\ \frac{1}{cosh\ x} \\coth\ x\ =\ \frac{1}{tanh\ x} tanh x = cosh xsinh xcsch x = sinh x1sech x = cosh x1coth x = tanh x1
然后额外补充一些性质:
sinhxisoddcoshxisevenwehavet∈Rthencosh2t−sinh2t=1sinh\ x\ is\ odd \\cosh\ x\ is\ even \\we\ have\ t\ \in\ R\ then\ \\cosh^{2}\ t\ -\ sinh^{2}\ t\ =\ 1 sinh x is oddcosh x is evenwe have t ∈ R then cosh2 t − sinh2 t = 1
其实最后一个结论的几何意义也就是等轴双曲线的一支,而且很显然焦点在xxx轴上的友半支或者是焦点在yyy轴的上半支,这就取决于你认为谁是横坐标谁是纵坐标了。
和差化积与积化和差公式
记得高中的时候老师就在课上提到过这些个公式,在大学有提到了这些个公式,高中时因为知道记了没用于是乎直接放弃,到了大学好像在上课时睡过去了导致再次错过
公式表述
1.首先是积化和差
sinα+sinβ=2sinα+β2cosα−β2sinα−sinβ=2cosα+β2sinα−β2cosα+cosβ=2cosα+β2cosα−β2cosα−cosβ=−2sinα+β2sinα−β2sin\ \alpha\ +\ sin\ \beta\ =\ 2sin\ \frac{\alpha+\beta}{2}cos\ \frac{\alpha-\beta}{2} \\sin\ \alpha\ -\ sin\ \beta\ =\ 2cos\ \frac{\alpha+\beta}{2}sin\ \frac{\alpha-\beta}{2} \\cos\ \alpha\ +\ cos\ \beta\ =\ 2cos\ \frac{\alpha+\beta}{2}cos\ \frac{\alpha-\beta}{2} \\cos\ \alpha\ -\ cos\ \beta\ =\ -2sin\ \frac{\alpha+\beta}{2}sin\ \frac{\alpha-\beta}{2} sin α + sin β = 2sin 2α+βcos 2α−βsin α − sin β = 2cos 2α+βsin 2α−βcos α + cos β = 2cos 2α+βcos 2α−βcos α − cos β = −2sin 2α+βsin 2α−β
2.然后是和差化积
sinαcosβ=sin(α+β)+sin(α−β)2cosαsinβ=sin(α+β)−sin(α−β)2cosαcosβ=cos(α+β)+cos(α−β)2sinαsinβ=−cos(α+β)−cos(α−β)2sin\ \alpha\ cos\ \beta\ =\ \frac{sin\ (\alpha+\beta)\ +\ sin\ (\alpha-\beta)}{2} \\cos\ \alpha\ sin\ \beta\ =\ \frac{sin\ (\alpha+\beta)\ -\ sin\ (\alpha-\beta)}{2} \\cos\ \alpha\ cos\ \beta\ =\ \frac{cos\ (\alpha+\beta)\ +\ cos\ (\alpha-\beta)}{2} \\sin\ \alpha\ sin\ \beta\ =\ -\frac{cos\ (\alpha+\beta)\ -\ cos\ (\alpha-\beta)}{2} sin α cos β = 2sin (α+β) + sin (α−β)cos α sin β = 2sin (α+β) − sin (α−β)cos α cos β = 2cos (α+β) + cos (α−β)sin α sin β = −2cos (α+β) − cos (α−β)
有一说一是真的难记
公式证明
当然真正证明其实构造扩展然后打开一边应该是很显然,虽然我还没有试过,1013不考证明,但我认为最重要的是怎么正着推出这个公式来避免在考场上忘记或者记错。
1.对于积化和差
我认为我们在记录积化和差时其实只要抓住一个比较重要的点,就是把两边的自变量尽量凑成差不多的:
α=α+β2+α−β2β=α+β2−α−β2\alpha\ =\ \frac{\alpha+\beta}{2}\ +\ \frac{\alpha-\beta}{2} \\ \beta\ =\ \frac{\alpha+\beta}{2}\ -\ \frac{\alpha-\beta}{2} α = 2α+β + 2α−ββ = 2α+β − 2α−β
然后我们拿第一个式子来练练手:
sinα+sinβ=sin(α+β2+α−β2)+sin(α+β2−α−β2)=2sinα+β2cosα−β2(Evidently)sin\ \alpha\ +\ sin\ \beta \\=sin\ (\frac{\alpha+\beta}{2}\ +\ \frac{\alpha-\beta}{2})\ +\ sin\ (\frac{\alpha+\beta}{2}\ -\ \frac{\alpha-\beta}{2}) \\=2sin\ \frac{\alpha+\beta}{2}cos\ \frac{\alpha-\beta}{2} \\(Evidently) sin α + sin β=sin (2α+β + 2α−β) + sin (2α+β − 2α−β)=2sin 2α+βcos 2α−β(Evidently)
2.对于和差化积
这个很是简单,我们只要写出三角函数的四个展开式然后倒一倒就可以了,本来觉得太麻烦了实在不想写,最后还是觉得写一遍加深印象,顺便当作是熟悉markdown用法了
sin(α+β)=sinαcosβ+sinβcosαsin(α−β)=sinαcosβ−sinβcosαcos(α+β)=cosαcosβ−sinβsinαocs(α−β)=cosαcosβ+sinβsinαsin (\alpha+\beta)\ =\ sin\ \alpha\cos\ \beta\ +\ sin\ \beta\ cos\ \alpha\\ sin (\alpha-\beta)\ =\ sin\ \alpha\cos\ \beta\ -\ sin\ \beta\ cos\ \alpha\\ cos (\alpha+\beta)\ =\ cos\ \alpha\cos\ \beta\ -\ sin\ \beta\ sin\ \alpha\\ ocs (\alpha-\beta)\ =\ cos\ \alpha\cos\ \beta\ +\ sin\ \beta\ sin\ \alpha\\ sin(α+β) = sin αcos β + sin β cos αsin(α−β) = sin αcos β − sin β cos αcos(α+β) = cos αcos β − sin β sin αocs(α−β) = cos αcos β + sin β sin α
前两个式子一加除以222就是我们给出的第一个公式
极限
某些重要定理(及其可能比较重要的极限)
Sandwich Theorem
定理表述
又被称为夹逼定理,也咩有什么好证明的,不过是要注意一下前提条件之类的,还有就是其实有的时候也不一定能想起来这个定理,于是还是把它写一遍,再用它证明一些极限,强化一下记忆。
FirstlywenotearangeIthatforf(x),g(x),h(x)arealldefinedinthisrangealso,inthisrange:g(x)≤f(x)≤h(x),(x∈Iandx≠a)thenforthevaluea∈Ilimx→ag(x)=Llimx→ah(x)=Lthen:limx→af(x)=LFirstly\ we \ note\ a\ range\ I\ that\ for\ f(x),g(x),h(x)\ are\ all\ defined\ in\ this\ range \\also,\ in\ this\ range: \\g(x)\le f(x) \le h(x),(x\in I\ and\ x\ne a) \\then\ for\ the\ value\ a\in I \\\lim_{x \rightarrow a}g(x)=L \\\lim_{x \rightarrow a}h(x)=L \\then: \\\lim_{x \rightarrow a}f(x)=L Firstly we note a range I that for f(x),g(x),h(x) are all defined in this rangealso, in this range:g(x)≤f(x)≤h(x),(x∈I and x=a)then for the value a∈Ix→alimg(x)=Lx→alimh(x)=Lthen:x→alimf(x)=L
定理的表述看起来有些复杂,不过具体使用时好像还是比较清楚的。
这里有一个坑需要注意的是我们在遇到某些两边不是≤\le≤的情况,只要能在所得点夹挤仍然可以是使用夹挤定理的,例子详见证明:limx→0sinx=0\lim_{x\rightarrow 0}sin\ x=0limx→0sin x=0
定理的证明
本来觉得这样一个简单的定理应该是很显然不用证明的,结果看到了WIKI最下面的内容,竟一时间没有理解,于是先在这里贴出证明,慢慢消化了。不过话说考试也不可能考证明。
定理的应用
见“一些三角函数的极限”
某些重要极限及其证明
写在最前
这一部分包括了一些憨憨博主(很菜)感觉到比较有趣的极限,主要是因为我太蒻,导致可能很多基本的东西对我来说就是天书,所以还是总结一下,反正到了大学也不会有人刷题了吧。不会吧不会吧。
e=limx→∞(1+1x)xe= \lim_{x\rightarrow \infty}(1+\frac{1}{x})^{x}e=limx→∞(1+x1)x
有一说一这个极限好像没法证明,因为他本来就是eee的定义啊,跳过跳过。
加一个另一种写法:
e=limx→∞(1+x)1xe=\lim_{x\rightarrow \infty}(1+x)^{\frac{1}{x}} e=x→∞lim(1+x)x1
对于新加的写法,在“solution to application”里已经有了解释,现在对第一种·标题里的写法我们再做一下解释:
asweknowthat:1∞=1however,iff(x)→1andg(x)→∞(f(x)>1)limf(x)g(x)=eas\ we\ know\ that: \\1^{\infty}=1 \\however,if\ f(x)\rightarrow 1\ and\ g(x)\rightarrow\infty\ (f(x)>1) \\\lim f(x)^{g(x)}=e as we know that:1∞=1however,if f(x)→1 and g(x)→∞ (f(x)>1)limf(x)g(x)=e
APPLICATION:
Calculatefollowinglimits:1.limx→∞(1−1x−1x2)x2.limx→∞(1+2x+3)xCalculate\ following\ limits: \\ 1.\lim_{x\rightarrow \infty}(1-\frac{1}{x}-\frac{1}{x^2})^x \\ 2.\lim_{x\rightarrow \infty}(1+\frac{2}{x+3})^x Calculate following limits:1.x→∞lim(1−x1−x21)x2.x→∞lim(1+x+32)x
一些三角函数的极限
limx→asinx=sina\lim_{x \rightarrow a}sin\ x=sin\ alimx→asin x=sin a
PROOF:
Actuallyitisnotasimplequestion,justasthenormalways,wefirstlyconsiderthespecialcases:whenx→0Firstlyasweknowfor0≤x≤π2:0<sinx≤1accordingtothesandwichtheorem:whenisbiggerthan0andapproch0,wecanknow:limx→0+sinx=0whenx<0:limx→0−sinx=limx→0−−sin(−x)=lim−x→0+−sin(−x)=0whenx≠0:firstlywerecallanequation:sinα−sinβ=2sinα−β2cosα+β2soitisobviouslythat:∣sinx−sina∣=∣2sinx−a2cosx+a2∣≤∣2sinx−a2∣actaullyherewefindausefulmethodinsolvinglimitproblems:UsingtheabsolutevaluetocreatSandwichTheorem:itisobviouslythat:−∣2sinx−a2∣≤sinx−sina≤∣2sinx−a2∣sobyapplySandwichTheorem:limx→asinx=sinasofromthetwocases,wefinallyprovethetheoremActually\ it\ is\ not\ a\ simple\ question,\ just\ as\ the\ normal\ ways,\ we\ firstly\ consider\ the\ special\ cases: \\when\ x\rightarrow 0 \\Firstly\ as\ we\ know\ for\ 0 \le x\le \frac{\pi}{2}: \\0< sin\ x\le1 \\according\ to\ the\ sandwich\ theorem: \\when\ is\ bigger\ than\ 0\ and\ approch\ 0,we\ can\ know: \\\lim_{x\rightarrow 0^{+}}sin\ x=0 \\when\ x<0: \\\lim_{x\rightarrow 0^{-}}sin\ x=\lim_{x\rightarrow 0^{-}}-sin\ (-x)=\lim_{-x\rightarrow 0^{+}}-sin\ (-x)=0 \\ \ \ \\when\ x\ne0: \\firstly\ we\ recall\ an\ equation: \\sin\ \alpha-sin\ \beta=2sin\ \frac{\alpha-\beta}{2}cos\ \frac{\alpha+\beta}{2} \\so\ it\ is\ obviously\ that: \\|sin\ x-sin\ a|= |2sin\ \frac{x-a}{2}cos\ \frac{x+a}{2}|\le|2sin\ \frac{x-a}{2}| \\actaully\ here\ we\ find\ a\ useful\ method\ in\ solving\ limit\ problems: \\Using\ the\ absolute \ value\ to\ creat\ Sandwich\ Theorem: \\it\ is\ obviously\ that: \\-|2sin\ \frac{x-a}{2}|\le sin\ x-sin\ a\le |2sin\ \frac{x-a}{2}| \\so\ by\ apply\ Sandwich\ Theorem: \\\lim_{x\rightarrow a}sin\ x=sin\ a \\ \ \ \ \\so\ from\ the\ two\ cases,we\ finally\ prove\ the\ theorem Actually it is not a simple question, just as the normal ways, we firstly consider the special cases:when x→0Firstly as we know for 0≤x≤2π:0<sin x≤1according to the sandwich theorem:when is bigger than 0 and approch 0,we can know:x→0+limsin x=0when x<0:x→0−limsin x=x→0−lim−sin (−x)=−x→0+lim−sin (−x)=0 when x=0:firstly we recall an equation:sin α−sin β=2sin 2α−βcos 2α+βso it is obviously that:∣sin x−sin a∣=∣2sin 2x−acos 2x+a∣≤∣2sin 2x−a∣actaully here we find a useful method in solving limit problems:Using the absolute value to creat Sandwich Theorem:it is obviously that:−∣2sin 2x−a∣≤sin x−sin a≤∣2sin 2x−a∣so by apply Sandwich Theorem:x→alimsin x=sin a so from the two cases,we finally prove the theorem
limx→0sinxx=1\lim_{x \rightarrow 0}\frac{sin\ x}{x}=1limx→0xsin x=1
PROOF:
AccordingtotheunknowproblemssotheproofhasbeenguguguPIGEONHEREAccording\ to\ the\ unknow\ problems\ so\ the\ proof\ has\ been\ gugugu \\PIGEON\ HERE According to the unknow problems so the proof has been guguguPIGEON HERE
APPLICATION:
Calculatethefollowinglimits:1.limx→0sin(sinx)x2.limx→∞2xsinsina2xCalculate\ the\ following\ limits: \\1. \lim_{x\rightarrow 0}\frac{sin(sin\ x)}{x} \\ 2. \lim_{x\rightarrow \infty} 2^{x}sin\frac{sin\ a}{2^{x}} Calculate the following limits:1.x→0limxsin(sin x)2.x→∞lim2xsin2xsin a
limx→0x2sin1x=0\lim_{x \rightarrow 0}x^{2}sin\ \frac{1}{x}=0limx→0x2sin x1=0
PROOF;
aswefirstlyseethisformulawecanbesuretouseSandwichTheorem:forthesin,cos...weneedtouseitsrangeandabsolutevaluetoconstructSandwichTheorem:when0<x≤π2∣x2sin1x∣≤x2→−x2≤x2sin1x≤x2andwhenx=0,bothx2and−x2equal0soitisobviouslythat:limx→0x2sin1x=0as\ we\ firstly\ see\ this\ formula\ we\ can\ be\ sure\ to\ use\ Sandwich\ Theorem : \\for\ the\ sin,cos...\ we\ need\ to\ use\ its\ range\ and\ absolute\ value\ to\ construct\ Sandwich\ Theorem: \\when\ 0<x\le \frac{\pi}{2} \\|x^2sin\ \frac{1}{x}|\le x^2 \\\rightarrow -x^2\le x^2sin\ \frac{1}{x}\le x^2 \\and\ when\ x=0,\ both\ x^2\ and\ -x^2\ equal\ 0 \\so\ it\ is\ obviously\ that: \\\lim_{x\rightarrow\ 0}x^2sin\ \frac{1}{x}=0 as we firstly see this formula we can be sure to use Sandwich Theorem:for the sin,cos... we need to use its range and absolute value to construct Sandwich Theorem:when 0<x≤2π∣x2sin x1∣≤x2→−x2≤x2sin x1≤x2and when x=0, both x2 and −x2 equal 0so it is obviously that:x→ 0limx2sin x1=0
solution to the APPLICATION of important limits
为什么要写这个呢,主要是因为把博客快写完之后才发现原来最重要的极限基本都默认的是这两个,所以就把网上比较经典的应用整理了下来,做一个总结,计算完之后可能会得出一些对于极限求法的思考吧(雾)
Atfirstweaddanewlawoflimitsbecauseitisusefulespecallyusefulwhencalculatinglimitsaboutelimx→af(x)=Aandlimx→ag(x)=Bso:limx→af(x)g(x)=ABAlsoaswetalkthisformoflimits,weneedtoknowthat:∞0≠1(notaprecisefoemula,butenoughtoshowwhatImean)actuallyit′seasytounderstandbecauseweknow:limx→∞(1+x)1x=esolutionhere:1.limx→∞(1−1x−1x2)x=limx→∞(x2−x−1x2)x=limx→∞(1+−(x+1)x)x=limx→∞(1+−(x+1)x)−xx+1×−x+1x2=limx→∞((1+−(x+1)x)−xx+1)−x+1x2=limx→∞(1+−(x+1)x)−xx+1×limx→∞−x+1x2=(e−1)0=1e2.limx→∞(1+2x+3)x=limx→∞(1+1x+32)x=limx→∞(1+1(x+3)2)x+32×2xx+3=limx→∞((1+1(x+3)2)x+32)limx→∞2xx+3=e2×limx→∞xx+3=e23.limx→0sin(sinx)x=limx→0(sin(sinx)sinx×sinxx)=limsinx→0sin(sinx)sinx×limx→0sinxx=limu→0sinuu×limx→0sinxx=1(whichu=sinx)4.limx→∞2xsinsina2x=limx→∞(sinsina2xsina2x×sina)=1×limx→∞sina=sinaKEY:trytousemultiplicationanddivision(betterthanaddandminus)tochangetheformulatotheimportantlimits\\At\ first\ we\ add\ a\ new\ law\ of\ limits\ because\ it\ is\ useful\ especally \ useful\ when\ calculating\ limits\ about\ e \\\lim_{x\rightarrow a}f(x)=A\ and\ \lim_{x\rightarrow a}g(x)=B \\so:\lim_{x\rightarrow a}f(x)^{g(x)}=A^B \\Also\ as\ we\ talk\ this\ form\ of\ limits,\ we\ need\ to\ know\ that: \\\infty ^{0}\ne1 \\(not\ a\ precise\ foemula,\ but\ enough\ to\ show\ what\ I mean) \\actually\ it's\ easy\ to\ understand\ because\ we\ know: \\ \lim_{x\rightarrow \infty}(1+x)^{\frac{1}{x}}=e \\ \ \ \ \\solution\ here: \\ 1.\lim_{x\rightarrow \infty}(1-\frac{1}{x}-\frac{1}{x^2})^x \\ =\lim_{x\rightarrow \infty} (\frac{x^2-x-1}{x^2})^x=\lim_{x \rightarrow \infty} (1+\frac{-(x+1)}{x})^x \\=\lim_{x\rightarrow \infty}(1+\frac{-(x+1)}{x})^{\frac{-x}{x+1}\times \frac{-x+1}{x^2}} \\=\lim_{x\rightarrow \infty}((1+\frac{-(x+1)}{x})^{\frac{-x}{x+1})^{\frac{-x+1}{x^2}}} \\=\lim_{x\rightarrow \infty}(1+\frac{-(x+1)}{x})^{\frac{-x}{x+1}\times \lim_{x\rightarrow \infty}\frac{-x+1}{x^2}} \\=(e^{-1})^0=\frac{1}{e} \\ \ \ \ \\ \ \ \ \ 2.\lim_{x\rightarrow \infty}(1+\frac{2}{x+3})^x \\=\lim_{x\rightarrow \infty}(1+\frac{1}{\frac{x+3}{2}})^x=\lim_{x\rightarrow \infty}(1+\frac{1}{\frac{(x+3)}{2}})^{\frac{x+3}{2}\times\frac{2x}{x+3}} \\=\lim_{x\rightarrow \infty}((1+\frac{1}{\frac{(x+3)}{2}})^{\frac{x+3}{2})^{\lim_{x\rightarrow\infty}\frac{2x}{x+3}}}=e^{2\times \lim_{x \rightarrow \infty}\frac{x}{x+3}}=e^2 \\ \ \ \ \\ \ \ \ 3.\lim_{x\rightarrow 0}\frac{sin(sin\ x)}{x} \\=\lim_{x\rightarrow 0}(\frac{sin(sin\ x)}{sin\ x}\times\frac{sin\ x}{x})=\lim_{sin\ x\rightarrow 0}\frac{sin(sin\ x)}{sin\ x}\times\lim_{x\rightarrow 0}{\frac{sin\ x}{x}} \\=\lim_{u\rightarrow 0}{\frac{sin\ u}{u}}\times\lim_{x\rightarrow 0}{\frac{sin\ x}{x}}=1 \\(which\ u=sin\ x) \\ \ \ \ \\\ \ \ 4.\lim_{x\rightarrow \infty}2^xsin\frac{sin\ a}{2^x} \\=\lim_{x\rightarrow \infty}(\frac{sin\frac{sin\ a}{2^x}}{\frac{sin\ a}{2^x}}\times sin\ a)=1\times \lim_{x\rightarrow\infty}sin\ a \\=sin\ a \\\ \ \ \ \\\ \ \ KEY:try\ to\ use\ multiplication\ and\ division(better\ than\ add\ and\ minus)to\ change\ the\ formula\ to\ the\ important\ limits At first we add a new law of limits because it is useful especally useful when calculating limits about ex→alimf(x)=A and x→alimg(x)=Bso:x→alimf(x)g(x)=ABAlso as we talk this form of limits, we need to know that:∞0=1(not a precise foemula, but enough to show what Imean)actually it′s easy to understand because we know:x→∞lim(1+x)x1=e solution here:1.x→∞lim(1−x1−x21)x=x→∞lim(x2x2−x−1)x=x→∞lim(1+x−(x+1))x=x→∞lim(1+x−(x+1))x+1−x×x2−x+1=x→∞lim((1+x−(x+1))x+1−x)x2−x+1=x→∞lim(1+x−(x+1))x+1−x×limx→∞x2−x+1=(e−1)0=e1 2.x→∞lim(1+x+32)x=x→∞lim(1+2x+31)x=x→∞lim(1+2(x+3)1)2x+3×x+32x=x→∞lim((1+2(x+3)1)2x+3)limx→∞x+32x=e2×limx→∞x+3x=e2 3.x→0limxsin(sin x)=x→0lim(sin xsin(sin x)×xsin x)=sin x→0limsin xsin(sin x)×x→0limxsin x=u→0limusin u×x→0limxsin x=1(which u=sin x) 4.x→∞lim2xsin2xsin a=x→∞lim(2xsin asin2xsin a×sin a)=1×x→∞limsin a=sin a KEY:try to use multiplication and division(better than add and minus)to change the formula to the important limits
导数
隐函数求导
想想自己高考压轴题好像是因为用了隐函数求导才扣了分,如果不扣是不是可以140+了呢?这么一说高中还不如不掌握。
高中的simple task
第一次听说隐函数求导是在高中的时候,如果没记错的话是老师在圆锥曲线中补充的内容,因为上课抛出了一个问题:怎么在短时间内快速地求出圆锥曲线在某一点的切线方程,尤其对于双曲线与椭圆来说。因为我们知道圆锥曲线的解析几何表达本质上只是一种xxx与yyy的对应关系,虽然它不是函数,但是还是可以求出yyy对于xxx的导数然后求出结果。为了让这篇博客看起来更长,我们先回忆一下三种最基本的圆锥曲线:
x2a2+y2b2=1x2a2−y2b2=1y2=2px\frac{x^{2}}{a^{2}}\ +\ \frac{y^{2}}{b^{2}}\ =\ 1 \\\frac{x^{2}}{a^{2}}\ -\ \frac{y^{2}}{b^{2}}\ =\ 1 \\y^{2}\ =\ 2px a2x2 + b2y2 = 1a2x2 − b2y2 = 1y2 = 2px
我们仅讨论高中时候遇到最多的情况,焦点在xxx轴上,现在我们来回忆一下实现思路:
把xxx与yyy分别放在等号的两边,然后抓住重点就是两边同时求导,并且我们注意到此时yyy是复合函数,这也是这种方法的关键所在,我们就愉快地求出了切线方程的通式:
y2=b2−a2b2x2simultaneousderivationofbothsides:2y×y′=−2a2b2xy′=−2a2xb2yy^{2}=b^{2}\ -\ \frac{a^{2}}{b^{2}}x^{2} \\simultaneous\ derivation\ of\ both\ sides: \\2y\ \times\ y^{'}=-2\frac{a^{2}}{b^{2}}x \\y^{'}=-2\frac{a^{2}x}{b^{2}y} y2=b2 − b2a2x2simultaneous derivation of both sides:2y × y′=−2b2a2xy′=−2b2ya2x
大学的hard task(雾)
粗略阅读Weiping的笔记,我们发现其实很大一部分仍然是两边同时求导,只不过可能会麻烦一些,基础思路都是把y′y^{'}y′单独表示出来,然后还有一类是要通过取对数变换这些操作完成,我们以一个经典题目为例,为什么说它经典,是因为从高中听到大学,老师讲了两年,但我仍未掌握,与其说没掌握,不如说没听过。
findthederivativeofthefollowingfunction:y=xxfind \ the\ derivative\ of\ the\ following\ function: \\y=x^{x} find the derivative of the following function:y=xx
我们先观察一下这种阴间结构没办法用最基本的法则表达出来,然后我们想到一个万金油思路,在高等代数中,碰到难以解决的高次项问题,我们的第一思路是取对数降次,所以我们选择自然对数,其实这是一种很重要的思想,取对数往往能将看似复杂的问题变得异常简单。要得开干!
y=xxnaturallogforbothsides:lny=xlnxcalculatethedirvativeofbothsides:y′×1y=1+lnxy′=y+ylnxorwenoteitas(thebetterandformerexpression):y′=xx(1+lnx)y=x^{x} \\natural\ log\ for\ both\ sides: \\ln\ y=xln\ x \\calculate\ the\ dirvative\ of\ both\ sides: \\y^{'}\times\frac{1}{y}=1\ +\ ln\ x \\y^{'}=y\ +\ yln\ x \\or\ we\ note\ it\ as\ (the\ better\ and\ former\ expression): \\y^{'}=x^{x}(1\ +\ ln\ x) y=xxnatural log for both sides:ln y=xln xcalculate the dirvative of both sides:y′×y1=1 + ln xy′=y + yln xor we note it as (the better and former expression):y′=xx(1 + ln x)
我们最后能化成只有自变量的是最好的,注意完美性与鲁棒性。值得一提的是其实作为一个水题自己却搁置了很久,然后第一遍上手还做错了,最主要的是右边少乘了一个yyy,只能说彩笔本质暴露无遗。
新的思考
正当我准备进行下一part的总结时,破天荒的头一次细看了一下note,居然发现了一些新的东西。原来之前自己的格局小的不行了,一直觉得隐函数求导只能用于基础法则不适用的函数,但对于某些复杂的函数也有奇效,我们先再看一道Weiping′snoteWeiping's \ noteWeiping′s note的例题,再来分析一下在大格局中它的用处。
findthedirvativeofthefollowingfunction:y=ex1−x21+x+x2find\ the\ dirvative\ of\ the\ following\ function: \\y=\frac{e^{x}\sqrt{1-x^{2}}}{1+x+x^{2}} find the dirvative of the following function:y=1+x+x2ex1−x2
当然对于头铁的人来说强上一点问题都没有,不过我们仔细观察一下这个函数的结构,存在自然对数,三个子函数之间都是用乘法与除法相连接起来。我们回忆一下高中数学的内容,什么运算符有加减运算法则而没有乘除运算法则呢?——原来是两边取对数。
格局打开
其实这又回到了我们刚刚总结过的问题,也就是在代数中碰到复杂的式子,尤其涉及到乘除运算时,取对数一定是我们的第一选择,再加上隐函数求导已经不是非法手段,所以可以随便取了。不说废话了,开始表演。
naturallogforbothsides:lny=ln(ex1−x21+x+x2)lny=lnex+ln(1−x2)−ln(1+x+x2)lny=x+12ln(1−x2)−ln(1+x+x2)findthedirvativeofbothsides:y′y=1+−2x2(1−x2)+1+2x1+x+x2y′=y(1+−2x2(1−x2)+1+2x1+x+x2)y′=ex1−x21+x+x2(1−x(1−x2)+1+2x1+x+x2)payattentiontothesimplestexpressionnatural\ log\ for\ both\ sides: \\ln\ y=ln\ (\frac{e^{x}\sqrt{1-x^{2}}}{1+x+x^{2}}) \\ln\ y=ln\ e^{x}\ +\ ln\ (\sqrt{1-x^{2}})\ -\ ln\ (1+x+x^{2}) \\ln\ y=x\ +\ \frac{1}{2}ln\ (1-x^{2})\ -\ ln\ (1+x+x^{2}) \\find\ the\ dirvative\ of\ both\ sides: \\\frac{y^{'}}{y}=1\ +\ \frac{-2x}{2(1-x^{2})}\ +\ \frac{1+2x}{1+x+x^{2}} \\y^{'}=y(1\ +\ \frac{-2x}{2(1-x^{2})}\ +\ \frac{1+2x}{1+x+x^{2}}) \\y^{'}=\frac{e^{x}\sqrt{1-x^{2}}}{1+x+x^{2}}(1\ -\ \frac{x}{(1-x^{2})}\ +\ \frac{1+2x}{1+x+x^{2}}) \\pay\ attention\ to\ the\ simplest\ expression natural log for both sides:ln y=ln (1+x+x2ex1−x2)ln y=ln ex + ln (1−x2) − ln (1+x+x2)ln y=x + 21ln (1−x2) − ln (1+x+x2)find the dirvative of both sides:yy′=1 + 2(1−x2)−2x + 1+x+x21+2xy′=y(1 + 2(1−x2)−2x + 1+x+x21+2x)y′=1+x+x2ex1−x2(1 − (1−x2)x + 1+x+x21+2x)pay attention to the simplest expression
至此我们回忆一下其实最重要的是其中蕴含的两种思想,一种是对于难处理的式子进行取对数降次,一种是两边同时求导等式仍然成立
感觉写了一堆就这两句话有用
反函数与原函数的导数关系
讨论之前我们先做一个假设:
nowwehavefunctionf(x)anditsinversefunctionf−1(x)now\ we\ have\ function\ f(x)\ and\ its\ inverse\ function\ f^{-1}(x) now we have function f(x) and its inverse function f−1(x)
然后回忆一下对于导数的两种表示方法:
f′(x)=df(x)dxf^{'}(x)=\frac{df(x)}{dx} f′(x)=dxdf(x)
所以不难想出对于它的反函数来说,导数应该是原函数导数的倒数 ,于是:
wenoteg(x)asf′(x)g′(x)=dg(x)dx=dg(x)df(g(x))=1f′(g(x))we\ note \ g(x)\ as\ f^{'}(x)\ \\g^{'}(x)=\frac{dg(x)}{dx}=\frac{dg(x)}{df(g(x))}=\frac{1}{f^{'}(g(x))} we note g(x) as f′(x) g′(x)=dxdg(x)=df(g(x))dg(x)=f′(g(x))1
其实好像这里不是很清晰的说明了这个问题,于是我们参考一下知乎与WeipingWeipingWeiping的笔记。然后我们发现他们的证明基本是大同小异,具体都是使用ChainRuleChain\ RuleChain Rule然后得到;
f′(g(x))×g′(x)=1f^{'}(g(x))\times g^{'}(x)=1 f′(g(x))×g′(x)=1
简略的记录一下具体得到的过程:
firstlyweknowthat:f(g(x))=xsofindthedirvativeoff(x):f(g(x))′=f′(g(x))×g′(x)=1firstly\ we\ know\ that: \\f(g(x) )= x \\so\ find\ the\ dirvative\ of\ f(x): \\f(g(x))^{'}=f^{'}(g(x))\times g^{'}(x)=1 firstly we know that:f(g(x))=xso find the dirvative of f(x):f(g(x))′=f′(g(x))×g′(x)=1
写到这里憨憨博主才意识到一个问题,原来自己一直没有理解这两个导数之间的关系,并不只是简单的倒数,而需要我们用牛莱两种表示方法重新写一下这个定理:
g′(x)=1f′(g(x))ordg(x)dx=1df(u)du∣u=g(x)g^{'}(x)=\frac{1}{f^{'}(g(x))} \\or \\\frac{dg(x)}{dx}=\frac{1}{\frac{df(u)}{du} \lvert _{u=g(x)}} g′(x)=f′(g(x))1ordxdg(x)=dudf(u)∣u=g(x)1
其实说到这里我们举个例子也就很显然了:
(ex)′=ex(lnx′)=1xnotrf(x)=exandusetheruleabove:lnx=g(x)=f−1(x)g′(x)=1elnx=1xwhichisevident(e^{x})^{'}=e^{x} \\(lnx^{'})=\frac{1}{x} \\notr\ f(x)=e^{x}\ and\ use\ the\ rule\ above: \\lnx=g(x)=f^{-1}(x) \\g^{'}(x)=\frac{1}{e^{lnx}}=\frac{1}{x} \\which\ is\ evident (ex)′=ex(lnx′)=x1notr f(x)=ex and use the rule above:lnx=g(x)=f−1(x)g′(x)=elnx1=x1which is evident
写在最后的就是一定要记住这个公式不是简单的倒数关系,每次不确定时就想想如上举的例子就行了。
利用导数进行估算
这一点也是在WeipingWeipingWeiping的note中发现的,可能上课确实讲过,不过因为证明过于绕,所以甚至都没记住结论,于是我们在最先先给出结论:
f(x)≈f(a)+f′(a)×(x−a)f(x)\approx f(a)+f^{'}(a)\times (x-a) f(x)≈f(a)+f′(a)×(x−a)
然后我们参考note,来思索一下它的证明:
firstlyweknow:f′(a)=limx→af(x)−f(a)x−aandweknowthatf′(a)isaconstnumber:limx→a(f(x)−f(a)x−a−f′(a))=0nowwenotethat:ϵ=f(x)−f(a)x−a−f′(a)so:limx→aϵ=0alsoweknowthat:f(x)=f(a)+f′(a)×(x−a)+ϵ×(x−a)bothϵand(x−a)aresosmallwhicharealmost0sowecansaythat:f(x)≈f(a)+f′(a)×(x−a)firstly\ we\ know: \\f^{'}(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} \\and\ we\ know\ that\ f^{'}(a)\ is\ a\ const\ number: \\\lim_{x\rightarrow a} (\frac{f(x)-f(a)}{x-a}-f^{'}(a))=0 \\now\ we\ note\ that: \\\epsilon = \frac{f(x)-f(a)}{x-a}-f^{'}(a) \\so: \\\lim_{x \rightarrow a} \epsilon = 0 \\also\ we\ know\ that: \\f(x) = f(a)+f^{'}(a)\times(x-a)+\epsilon \times(x-a) \\both\ \epsilon \ and\ (x-a)\ are \ so\ small\ which\ are\ almost\ 0 \\so\ we\ can\ say\ that: \\f(x)\approx f(a)+f^{'}(a)\times (x-a) firstly we know:f′(a)=x→alimx−af(x)−f(a)and we know that f′(a) is a const number:x→alim(x−af(x)−f(a)−f′(a))=0now we note that:ϵ=x−af(x)−f(a)−f′(a)so:x→alimϵ=0also we know that:f(x)=f(a)+f′(a)×(x−a)+ϵ×(x−a)both ϵ and (x−a) are so small which are almost 0so we can say that:f(x)≈f(a)+f′(a)×(x−a)
我们在看完了整个证明过程后,其实发现与其说是证明不如说是在求导过程中意外发现的一个结论,这里需要注意的是aaa与f(x)f(x)f(x)的选取,是一个会出现坑的地方,例如:
Question:1.02≈?Question: \sqrt{1.02} \approx? Question:1.02≈?
其实我们可以有很多种选取方法,不过note中提到了一种我认为最简单的方法:
wenotethat:f(x)=1+xanda=0so:1.02=f(0.02)f(0.02)≈f(0)+f′(0)×(0.02−0)=1.01we\ note\ that: \\f(x)=\sqrt{1+x}\ and\ a=0 \\so: \\\sqrt{1.02}=f(0.02) \\f(0.02)\approx f(0)+f^{'}(0)\times(0.02-0)=1.01 we note that:f(x)=1+x and a=0so:1.02=f(0.02)f(0.02)≈f(0)+f′(0)×(0.02−0)=1.01
看到这里我们不禁想起了这样一个让我永远无法忘记的问题:
weknowthat:a=2ln1.01andb=ln1.02andc=1.04−1findthebiggestandsmallestnumberofthreewe\ know\ that: \\a=2ln1.01\ and\ b=ln1.02\ and\ c=\sqrt{1.04}-1 \\find \ the\ biggest\ and\ smallest\ number\ of\ three we know that:a=2ln1.01 and b=ln1.02 and c=1.04−1find the biggest and smallest number of three
来自2021普通高等学校招生全国统一考试数学乙卷12题,虽然说当时是想出了严格正解,但四个月后坐在HKUST的图书馆里,我早就忘记了如何切线不等式倒出了结果。(雾)
itisobviouslythat:a=ln1.012=ln1.0201>ln1.02→a>bsoweneedtogettheapproximatevalueofc:wenotethat:f(x)=1+xanda=0so:1.04=f(0.04)f(0.04)≈f(0)+f′(0)×(0.04−0)=1.02c≈0.02wenotethat:g(x)=ln(1+x)anda=0ln(1.02)=g(0.02)g(0.02)≈g(0)+g′(0)×(0.02−0)=0.02it\ is\ obviously\ that: \\a=ln1.01^{2}=ln1.0201>ln1.02\ \rightarrow \ a>b \\so\ we\ need\ to\ get\ the\ approximate\ value\ of\ c: \\we\ note\ that: \\f(x)=\sqrt{1+x}\ and\ a=0 \\so: \\\sqrt{1.04}=f(0.04) \\f(0.04)\approx f(0)+f^{'}(0)\times(0.04-0)=1.02 \\c\approx0.02 \\\\we\ note\ that: \\g(x)=ln(1+x)\ and\ a=0 \\ln(1.02) = g(0.02) \\g(0.02)\approx g(0)+g^{'}(0)\times(0.02-0)=0.02 it is obviously that:a=ln1.012=ln1.0201>ln1.02 → a>bso we need to get the approximate value of c:we note that:f(x)=1+x and a=0so:1.04=f(0.04)f(0.04)≈f(0)+f′(0)×(0.04−0)=1.02c≈0.02we note that:g(x)=ln(1+x) and a=0ln(1.02)=g(0.02)g(0.02)≈g(0)+g′(0)×(0.02−0)=0.02
于是就麻了
其实很简单的一个道理就是因为取值的不同会导致可能是不足估计值也可能是过剩估计值,其实也就图一乐,真正比大小还得看切线不等式。于是激起了憨憨博主的好胜心。
告辞了,哪天闲了再填坑
与绝对值有关
Y1S1好像这一部分的内容都和note有关,都是从note上扒下来的,主要是这样一个问题,参数带绝对值时求导所得和不带绝对值时求导所得是否相同?
不过又说起来带绝对值了以后可导不可导都两说。那我们考虑一个可导的例子:
y=ln∣x∣y=ln\vert x\vert y=ln∣x∣
反正在代数中,带有绝对值的问题我们从来都是分类讨论的,于是尝试一下:
whenx>0:evidently:y′=1xwhenx<0y=ln(−x)y′=1−x×(−1)=1xsowecansay:(ln∣x∣)′=1xwhen\ x>0: \\evidently:y^{'}=\frac{1}{x} \\when\ x<0 \\y=ln(-x) \\y^{'}=\frac{1}{-x}\times(-1)=\frac{1}{x} \\so\ we\ can\ say: \\(ln\vert x \vert)^{'}=\frac{1}{x} when x>0:evidently:y′=x1when x<0y=ln(−x)y′=−x1×(−1)=x1so we can say:(ln∣x∣)′=x1
大致就是在定义域\可导范围内分类讨论就vans了,不过说到这里还是记得判断一下可导不可导比较OK。
local maximum and absolute maximum
暂时不考先鸽掉
Tricky Question
Problem1:asweknowf(x)=xxandg(x)isitsinversefunctionfindg′(4)Problem2:findthedivrativeoftan−1(cos6x)Problem3:asweknowf(x)=∣x∣tan∣x∣32,findf′(0)Problem4:asweknowf(x)=4x−2xxfindlimx→0f(x)andwithoutLobita′sRuleProblem1:as\ we\ know\ f(x)=x^x\ and\ g(x)\ is\ its\ inverse\ function \\find \ g'(4) \\\ \ \ \ \\Problem2:find\ the\ divrative\ of\ tan^{-1}(cos\ 6x) \\ \ \ \ \ \\Problem3:as\ we\ know\ f(x)=\sqrt{|x|}\ tan\ |x|^{\frac{3}{2}}, \\find\ f'(0) \\\ \ \ \ \ \\Problem4:as\ we\ know\ f(x)=\frac{4^x-2^x}{x} \\find\ \lim_{x\rightarrow 0}f(x)\ and \ without\ Lobita's\ Rule Problem1:as we know f(x)=xx and g(x) is its inverse functionfind g′(4) Problem2:find the divrative of tan−1(cos 6x) Problem3:as we know f(x)=∣x∣ tan ∣x∣23,find f′(0) Problem4:as we know f(x)=x4x−2xfind x→0limf(x) and without Lobita′s Rule
Solution to Tricky Question
Sol.1firstlylet′sreviewhowtofindderivativeofaninversefunction:writeintwodifferentgiants′ways(supposeg(x)isaninversefunctionoff(x)):g′(x)=1f′(g(x))dg(x)dx=1df(u)du∣u=g(x)Withitisalmostimpossibletofindg(x),wejustconsiderf′(x)inonepointaccordingtoourformerrecord,weknowthat:f′(x)=xx(1+lnx)accordingtotheconnectionofinversefunctionandoriginalfunctionf(g(x))=x:f(g(4))=4→g(4)=2sog′(4)=1f′(2)=14+4ln2Sol.2AlsoaboutinversefunctionandoriginalfunctionandImadeamistakeatthefirsttimeastheformulainSol.1,firstly:notef(x)=tan−1(cos6x)andu=cos6xsof′(x)=f′(u)×u′f′(u)=11cos2(tanu)=11+u2u′=−6sin6xsof′(x)=−6sin6x×11+u2=−6sin6x1+cos26xSol.3Actuallyitistrickybecauseitisreallyeasytomakemistakeonthisproblembecauseifweinsistfindingthedervativeoff(x)(whenx≥0):wemayignorethatforx,theslopeofitstagentline(whenx=0)isinfinity,whichmeansitsdervativewhenx=0doesn′texist.ItisasaREMINDERwhenweuserulesofdervative,wemustfirstjudgeifitsdervativeexist.soweapplythedefinition:f′(0)=limx→0f(x)−f(0)x−0anddivideitintotwocases:case1:limx→0+f(x)−f(0)x−0=limx→0+xtanx32x=limx→0+xsinx32cosx32x=limx→0+sinx32x32×limx→0+xcosx32=1case2:limx→0−f(x)−f(0)x−0=limx→0−−xtan(−x)32x=1(asthesameway)Sol.4limx→0f(x)=limx→04x−2xx=limx→02x(2x−1)x=1×limx→02x−20x−0=1×(2x)′∣x=0=1×2xln2∣x=0=ln2wemayconcludefromProblem4thatlimitsanddervativeshavecloseconnectionwemayusedefinitiontofindadervativeiftherulesarenotfitwemaychangelimitstodervativesifit′snotclearforcalculatinglimits(00,∞∞,etc.)Sol.1 \\firstly\ let's\ review\ how\ to\ find\ derivative\ of\ an\ inverse\ function: \\write\ in\ two\ different\ giants'\ ways(suppose\ g(x)\ is\ an\ inverse\ function\ of\ f(x)): \\g'(x)=\frac{1}{f'(g(x))} \\\frac{dg(x)}{dx}=\frac{1}{\frac{df(u)}{du}|_{u=g(x)}} \\With\ it\ is\ almost\ impossible\ to\ find\ g(x),\ we\ just\ consider\ f'(x)\ in\ one\ point \\according\ to\ our\ former\ record,\ we\ know\ that: \\f^{'}(x)=x^{x}(1\ +\ ln\ x) \\according\ to\ the\ connection\ of\ inverse\ function\ and\ original\ function \\f(g(x))=x: \\f(g(4))=4\rightarrow g(4)=2 \\so\ g'(4)=\frac{1}{f'(2)}=\frac{1}{4+4ln2} \\ \ \ \ \\Sol.2 \\Also\ about \ inverse\ function\ and\ original\ function\ and\ I\ made\ a\ mistake\ at\ the\ first\ time \\as\ the\ formula\ in\ Sol.1,firstly: \\note\ f(x)=tan^{-1}(cos\ 6x)\ and\ u=cos\ 6x \\so\ f'(x)=f'(u)\times u' \\f'(u)=\frac{1}{\frac{1}{cos^2(tan\ u)}}=\frac{1}{1+u^2} \\u'=-6sin\ 6x \\so\ f'(x)=-6sin\ 6x\times \frac{1}{1+u^2}=\frac{-6sin\ 6x}{1+cos^2\ 6x} \\\ \ \ \ \\Sol.3 \\Actually\ it\ is\ tricky\ because\ it\ is\ really\ easy\ to\ make\ mistake\ on\ this\ problem \\because\ if\ we\ insist \ finding\ the\ dervative\ of\ f(x)\ (when\ x\ge 0): \\we\ may\ ignore\ that\ for\ \sqrt{x},the\ slope\ of\ its\ tagent\ line(when\ x=0)\ is\ infinity, \\which\ means\ its\ dervative\ when\ x=0\ doesn't\ exist. \\It\ is\ as\ a\ REMINDER\ when\ we\ use\ rules\ of\ dervative,\ we\ must\ first\ judge\ if\ its\ dervative\ exist. \\so\ we\ apply\ the\ definition: \\f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0} \\and\ divide\ it\ into\ two\ cases: \\case1: \\\lim_{x\rightarrow 0^+}\frac{f(x)-f(0)}{x-0} \\=\lim_{x\rightarrow 0^+}\frac{\sqrt{x}\ tan\ x^{\frac{3}{2}}}{x}=\lim_{x\rightarrow 0^+}\sqrt{x}\frac{sin\ x^\frac{3}{2}}{cos\ x^\frac{3}{2}x} \\=\lim_{x\rightarrow 0^+}\frac{sin\ x^\frac{3}{2}}{x^\frac{3}{2}}\times \lim_{x\rightarrow 0^+}\frac{x}{cos\ x^\frac{3}{2}}=1 \\case2: \\\lim_{x\rightarrow 0^-}\frac{f(x)-f(0)}{x-0} \\=\lim_{x\rightarrow 0^-}\frac{\sqrt{-x}\ tan\ (-x)^{\frac{3}{2}}}{x}=1(as\ the\ same\ way) \\ \ \ \ \ \\Sol.4 \\\lim_{x\rightarrow 0}f(x) \\=\lim_{x\rightarrow 0}\frac{4^x-2^x}{x}=\lim_{x\rightarrow 0}\frac{2^x(2^x-1)}{x} \\=1\times\lim_{x\rightarrow 0}\frac{2^x-2^0}{x-0}=1\times (2^x)'|_{x=0} \\=1\times2^xln\ 2|_{x=0}=ln\ 2 \\we\ may\ conclude\ from\ Problem\ 4\ that\ limits\ and\ dervatives\ have\ close\ connection \\we\ may\ use\ definition\ to\ find\ a\ dervative\ if\ the\ rules\ are\ not\ fit \\we\ may\ change\ limits\ to\ dervatives\ if\ it's\ not\ clear\ for\ calculating\ limits(\frac{0}{0},\frac{\infty}{\infty},etc.) Sol.1firstly let′s review how to find derivative of an inverse function:write in two different giants′ ways(suppose g(x) is an inverse function of f(x)):g′(x)=f′(g(x))1dxdg(x)=dudf(u)∣u=g(x)1With it is almost impossible to find g(x), we just consider f′(x) in one pointaccording to our former record, we know that:f′(x)=xx(1 + ln x)according to the connection of inverse function and original functionf(g(x))=x:f(g(4))=4→g(4)=2so g′(4)=f′(2)1=4+4ln21 Sol.2Also about inverse function and original function and I made a mistake at the first timeas the formula in Sol.1,firstly:note f(x)=tan−1(cos 6x) and u=cos 6xso f′(x)=f′(u)×u′f′(u)=cos2(tan u)11=1+u21u′=−6sin 6xso f′(x)=−6sin 6x×1+u21=1+cos2 6x−6sin 6x Sol.3Actually it is tricky because it is really easy to make mistake on this problembecause if we insist finding the dervative of f(x) (when x≥0):we may ignore that for x,the slope of its tagent line(when x=0) is infinity,which means its dervative when x=0 doesn′t exist.It is as a REMINDER when we use rules of dervative, we must first judge if its dervative exist.so we apply the definition:f′(0)=x→0limx−0f(x)−f(0)and divide it into two cases:case1:x→0+limx−0f(x)−f(0)=x→0+limxx tan x23=x→0+limxcos x23xsin x23=x→0+limx23sin x23×x→0+limcos x23x=1case2:x→0−limx−0f(x)−f(0)=x→0−limx−x tan (−x)23=1(as the same way) Sol.4x→0limf(x)=x→0limx4x−2x=x→0limx2x(2x−1)=1×x→0limx−02x−20=1×(2x)′∣x=0=1×2xln 2∣x=0=ln 2we may conclude from Problem 4 that limits and dervatives have close connectionwe may use definition to find a dervative if the rules are not fitwe may change limits to dervatives if it′s not clear for calculating limits(00,∞∞,etc.)
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