上海财经大学英语高数课件02

Chapter2Derivativesupdownreturnend2.6Implicitdifferentiation1)Explicitfunction:Thefunctionwhichcanbedescribedbyexpressingonevariableexplicitlyintermsofanothervariable(othervariables)aregenerallycalledexplicitfunction---forexample,y=xtanx,ory=[1+x2+x3]1/2,oringeneraly=f(x).2)Implicitfunction:Thefunctionswhicharedefinedimplicitlybyarelationbetweenvariables--xandy--aregenerallycalledimplicitfunctions---suchasx2+y2=4,or7sin(xy)=x2+y3or,ingeneralF(x,y)=0.Ify=f(x)satisfiesF(x,f(x))=0onanintervalI,wesayf(x)isafunctiondefinedonIimplicitlybyF(x,y)=0,orimplicitfunctiondefinedbyF(x,y)=0.updownreturnend3)DerivativesofimplicitfunctionSupposey=f(x)isanimplicitfunctiondefinedbysin(xy)=x2+y3.Thensin[xf(x)]=x2+[f(x)]3.Fromtheequation,wecanfindthederivativeoff(x)eventhoughwehavenotgottentheexpressionoff(x).Fortunatelyitisnotnecessarytosolvetheequationforyintermsofxtofindthederivative.Wewillusethemethodcalledimplicitdifferentiationtofindthederivative.Differentiatingbothsidesoftheequation,weobtainthat[f(x)+xf'(x)]cos[xf(x)]=2x+3[f(x)]2f'(x).Then)cos(32)cos()](cos[)](3[2)]([cos)('22xyxyxxyyxxfxxfx-xxff(x)xfupdownreturnendExample(a)Ifx3+y3=27,find.dxdy(b)Findtheequationofthetangenttothecurvex3+y3=28atpoint(1,3).updownreturnendExample(a)Ifx3+y3=6xy,findy'.(b)FindtheequationofthetangenttothefoliumofDescartesx3+y3=6xyatpoint(3,3).updownreturnendOrthogonal:TwocurvesarecalledOrthogonal,ifateachpointofintersectiontheirtangentlinesareperpendicular.Iftwofamiliesofcurvessatisfythateverycurveinonefamilyisorthogonaltoeverycurveintheanotherfamily,thenwesaythetwofamiliesofcurvesareorthogonaltrajectoresofeachother.ExampleTheequationsxy=c(c0)representsafamilyofhyperbolas.AndtheTheequationsx2-y2=k(k0)representsanotherfamilyofhyperbolaswithasymptotesy=x.Thenthetwofamiliesofcurvesaretrajectoresofeachother.updownreturnendDerivativef'(x)ofdifferentiablefunctionf(x)isalsoafunction.Iff'(x)isdifferentiable,thenwehave[f'(x)]'.Wewilldenoteitbyf''(x),i.e.,f''(x)=[f'(x)]'.Thenewfunctionf''(x)iscalledthesecondderivativeoff(x).Ify=f(x),wealsocanuseothernotations:Similarlyf'''(x)=[f''(x)]'iscalledthethirdderivativeoff(x),and)()()())(('')(''222222xfDxfDdxxfddxxdfdxddxydyxfx2.7Higherderivatives)()()())((''')('''33332233xfDxfDdxxfddxxfddxddxydyxfxupdownreturnendAndwecandefinef''''(x)=[f'''(x)]'.Fromnowoninsteadofusingf''''(x)weusef(4)(x)torepresentf''''(x).Ingeneral,wedefinef(n)(x)=[f(n-1)(x)]',whichiscalledthenthderivativeoff(x).Wealsoliketousethefollowingnotations,ify=f(x),ExampleIfy=x4-3x2+6x+9,findy',y'',y''',y(4).)()()())(()(11)()(xfDxfDdxxfddxxfddxddxydyxfnxnnnnnnnnnupdownreturnendExampleIff(x)=,findf(n)(x).x1ExampleIff(x)=sinx,g(x)=cosx,findf(n)(x)andg(n)(x).ExampleFindy'',ifx4+y3=x-y.2.8Relatedrates(omitted)2.9Differentials,LinearandQuadraticApproximationsDefinition:Letx=x-x0,f(x)=f(x)-f(x0).IfthereexistsaconstantA(x0)whichisindependentofxandxsuchthatf(x)=A(x0)x+B(x,x0)whereB(x,x0)satisfies.ThenAxiscalleddifferentialoff(x)atx0.GenerallyAxisdenotedbydf(x)|x=x0=A(x0)x.Replacingx0byx,thedifferentialisdenotedbydf(x)anddf(x)=A(x)x.0x)xB(x,lim00xupdownreturnendProof:Fromthedefinition,Corollary:Ifthedifferentialoff(x)isdf(x)=A(x)x,thenf(x)isdifferentiableandA(x)=f'(x).).(),()(lim)()(lim)('0xAtxtBtxAxtxftfxftxtCorollary:(a)Iff(x)=x,thendx=df(x)=x.(b)Iff(x)isdifferentiable,thendifferentialoff(x)existsanddf(x)=f'(x)dx.updownreturnendExample(a)Finddy,ify=x3+5x4.(b)Findthevalueofdywhenx=2anddx=0.1.Solution:Geometricmeaningofdifferentialoff(x),df(x)=QSf(x)=RSxoxyPtSRQdx=xdyy=f(x)Asx=dxisverysmall,y=dy,i.e.,f(t)-f(x)f'(x)t.updownreturnendExampleUsedifferentialstofindanapproximate(65)1/3.Fromdefinitionofthedifferential,wecaneasilygetIff(x)isdifferentiableatx=a,andxisveryclosedtoa,thenf(x)f(a)+f'(a)(x-a).TheapproximationiscalledLinearapproximationortangentlineapproximationoff(x)ata.AndfunctionL(x)=f(a)+f'(a)(x-a)iscalledthelinearizationoff(x)ata.updownreturnendExampleFindthelinearizationofthefunctionf(x)=(x+3)1/2andapproximationsthenumbers(3.98)1/2and(4.05)1/2.updownreturnendQuadraticapproximationtof(x)nearx=a:Supposef(x)isafunctionwhichthesecondderivativef''(a)exists.P(x)=A+Bx+Cx2istheparabolawhichsatisfiesP(a)=f(a),P'(a)=f'(a),andP''(a)=f''(a).Asxisveryclosedtoa,theP(x)iscalledQuadraticapproximationtof(x)neara.Corolary:SupposeP(x)=A+Bx+Cx2istheQuadraticapproximationtof(x)neara.ThenP(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)2/2.IfP(x)isthequadraticapproximationtof(x)nearx=a,thenasxisveryclosedtoa,P(x)f(x).Thatisf(x)f(a)+f'(a)(x-a)+f''(a)(x-a)1/2/2.updownreturnendExampleFindthequadraticapproximationtof(x)=cosxnear0.updownreturnendExampleFindthequadraticapproximationtof(x)=(x+3)1/2nearx=1.updownreturnendThemethodistogiveawaytogetaapproximationtoarootofanequation.2.10Newton’smethod(tobeomitted)Supposef(x)isdefinedon[a,b],f'(x)doesnotvalue0.Letx0[a,b],f(a)f(b)0.Andx1=x0-,x2=x1-.Keepingrepeatingtheprocess(xn=xn-1-),weobtainasequenceofapproximationsx1,x2,...,xn,......If,thenristherootoftheequationf(x)=0.)(xf')f(x00)(xf')