二導(dǎo)數(shù)及其應(yīng)用ppt課件

1、二二.導(dǎo)數(shù)及其應(yīng)用導(dǎo)數(shù)及其應(yīng)用求求 導(dǎo)導(dǎo) 法法 那么那么求求 導(dǎo)導(dǎo) 法法 那么那么根本公式根本公式根本公式根本公式導(dǎo)導(dǎo)數(shù)數(shù)導(dǎo)導(dǎo)數(shù)數(shù)xyxD DD DD D0lim導(dǎo)導(dǎo)數(shù)數(shù)導(dǎo)導(dǎo)數(shù)數(shù)xyxD DD DD D0lim關(guān)關(guān) 系系)( xodyydxydyydxdyD D+ += =D D = = = =關(guān)關(guān) 系系)( xodyydxydyydxdyD D+ += =D D = = = =高階導(dǎo)數(shù)高階導(dǎo)數(shù)高階導(dǎo)數(shù)高階導(dǎo)數(shù)高階微分高階微分高階微分高階微分微微 分分微微 分分xydyD D = =微微 分分微微 分分xydyD D = =洛必達(dá)法那么洛必達(dá)法那么洛必達(dá)法那么洛必達(dá)法那么Rolle定理定理Ro

2、lle定理定理LagrangeLagrange中值中值定理定理LagrangeLagrange中值中值定理定理常用常用的的泰勒公式泰勒公式常用常用的的泰勒公式泰勒公式型型00,1 ,0 型型 - - 型型 0型型00型型 型型00型型 CauchyCauchy中值定理中值定理CauchyCauchy中值定理中值定理TaylorTaylor中值定理中值定理TaylorTaylor中值定理中值定理xxF= =)()()(bfaf= =0= =ngfgf1= =fgfggf1111- -= =- -取對(duì)數(shù)取對(duì)數(shù)令令gfy = =單調(diào)性單調(diào)性, ,極值與最值極值與最值, ,凹凸性凹凸性, ,拐點(diǎn)拐點(diǎn),

3、 ,函數(shù)函數(shù)圖形的描繪圖形的描繪; ;曲率曲率; ;求根方法求根方法. .單調(diào)性單調(diào)性, ,極值與最值極值與最值, ,凹凸性凹凸性, ,拐點(diǎn)拐點(diǎn), ,函數(shù)函數(shù)圖形的描繪圖形的描繪; ;曲率曲率; ;求根方法求根方法. .導(dǎo)數(shù)的應(yīng)用導(dǎo)數(shù)的應(yīng)用單調(diào)性單調(diào)性, ,極值與最值極值與最值, ,凹凸性凹凸性, ,拐點(diǎn)拐點(diǎn), ,函數(shù)函數(shù)圖形的描繪圖形的描繪; ;曲率曲率; ;求根方法求根方法. .單調(diào)性單調(diào)性, ,極值與最值極值與最值, ,凹凸性凹凸性, ,拐點(diǎn)拐點(diǎn), ,函數(shù)函數(shù)圖形的描繪圖形的描繪; ;曲率曲率; ;求根方法求根方法. .導(dǎo)數(shù)的應(yīng)用導(dǎo)數(shù)的應(yīng)用1.導(dǎo)數(shù)定義的幾種等價(jià)形式導(dǎo)數(shù)定義的幾種等價(jià)形式

4、2.求導(dǎo)方法：求導(dǎo)方法：1用定義用定義2先化簡(jiǎn)再求導(dǎo)先化簡(jiǎn)再求導(dǎo)3將乘除變加減再求導(dǎo)將乘除變加減再求導(dǎo)4冪指函數(shù)求導(dǎo)法冪指函數(shù)求導(dǎo)法5隱函數(shù)求導(dǎo)法隱函數(shù)求導(dǎo)法6參數(shù)方程所確定的函數(shù)的導(dǎo)數(shù)參數(shù)方程所確定的函數(shù)的導(dǎo)數(shù)7籠統(tǒng)函數(shù)的高階導(dǎo)數(shù)籠統(tǒng)函數(shù)的高階導(dǎo)數(shù)3.導(dǎo)數(shù)的應(yīng)用導(dǎo)數(shù)的應(yīng)用 1中值定理中值定理 2函數(shù)的單調(diào)性與凹凸性函數(shù)的單調(diào)性與凹凸性 3極值應(yīng)用問題極值應(yīng)用問題1.導(dǎo)數(shù)定義的幾種等價(jià)形式導(dǎo)數(shù)定義的幾種等價(jià)形式0000000000000000000( )()(1)()lim()()()lim( )()(2)()lim( )()(3)()lim()( )limxhxxxxf xxf xf xxf

5、 xhf xf xhf xf xf xxxf xxf xf xxf xf xxx+ +- -= =+ +- -= =- -= =- - - -= =- - - -= =000( )(0)(0)lim( )(0)(0)lim( )(0)(0)limxhxfxffxf hffhf xffx- -= =- -= =- -= =00000000()()()lim()()()limhhf xhf xfxhf xhf xfxh- -+ +- -+ +-+-= =+-+-= =201sin,0( )0,10( )li( 351.8)m( )xxxf xxxf xf xP = = 求求例例指指導(dǎo)導(dǎo)例例并并討討

6、論論的的存存在在性性. .0112 sincos,0( )0,0 xxxf xxxx - - = = 時(shí)時(shí)，由由導(dǎo)導(dǎo)數(shù)數(shù)公公解解式式得得：2001sin0( )(0)(0)limlim0 xxxf xfxfxx- - - - - -= = = =00( )(0)00(0)limlim0 xxf xffxx+ +- -+ +- - -= = = =(0)0f=2.求導(dǎo)方法求導(dǎo)方法 1用定義用定義112 sincos,0( )0,0 xxf xxxx - - = = 故故0011(00)lim( )lim(2 sincos)xxff xxxx- - - -= = =- - 0lim( )xf x不

7、存在.不存在.0lim( )(0)xf xf注意：不要出現(xiàn)錯(cuò)誤：注意：不要出現(xiàn)錯(cuò)誤：不存在，不存在不存在，不存在0(0)lim( )xff x事實(shí)上，的存在性與極限事實(shí)上，的存在性與極限的存在性無關(guān).的存在性無關(guān).22,225,11( )11,1axf xxxaxPbx = = =+ + + + 例例 （指指導(dǎo)導(dǎo)求求 , ,b b使使在在處處改改）連連續(xù)續(xù)且且可可導(dǎo)導(dǎo). .2112(10)lim11(10)lim()xxfxfaxbab- -+ +- -= = =+ + += =+ += =+ +解解(1( )1,1(1)f xxabf= =+ +- - - -= =- -= =在在連連續(xù)續(xù)

8、則則21121( )(1)1(1)limlim11xxf xfxfxx- - - -+ +=-0( )02 212 2lim1(1)xxx- - - = = = - -+ +11( )(1)1(1)limlim11xxf xfaxbfxx+ +- -+ +- -+ +- -= = =- - -由由11lim1xaxaax+ +- -=- -( )1f xx = =在在可可導(dǎo)導(dǎo)11( )a= = - -代代入入得得b=2用定義求導(dǎo)適用于三種情形：用定義求導(dǎo)適用于三種情形：(1)分段函數(shù)包括絕對(duì)值函數(shù)在分段點(diǎn)處的導(dǎo)數(shù)分段函數(shù)包括絕對(duì)值函數(shù)在分段點(diǎn)處的導(dǎo)數(shù)(2)籠統(tǒng)函數(shù)或表達(dá)式較復(fù)雜的顯函數(shù)在某點(diǎn)的

9、導(dǎo)數(shù)籠統(tǒng)函數(shù)或表達(dá)式較復(fù)雜的顯函數(shù)在某點(diǎn)的導(dǎo)數(shù)(3) 與導(dǎo)數(shù)有關(guān)的證明題與導(dǎo)數(shù)有關(guān)的證明題11( )1lim3,3(1).( )372.1)1xxf xxffPx- -= = = 設(shè)設(shè)在在連連續(xù)續(xù)且且求求例例指指導(dǎo)導(dǎo)例例解解由由111lim3,lim( -1)0( )xxxxf x- -= = =且且1lim( )(1)0()xf xf= = =得得反反證證11( )(1)11(1)limlim113( )xxf xffxxf x- -= = = =- - -3( )sin3 ,(0).2xf xexf = = 設(shè)設(shè)求求332311( )sin33cos310.,30 xxxfxexxxxe

10、+ += = =用用此此法法不不能能求求出出在在的的導(dǎo)導(dǎo)數(shù)數(shù)解解0( )(0)(0)lim2xf xffx- -= =解解30sin3limxxexx= =30sin333limxxexx= =3( )(1)(23)(100),(3)f xx xxxf- - = =- - - 求求3( )(3)(3)lim3xf xffx- -= =- -解解3(1)(2)(3)(100)0lim3xx xxxxx- - - - - -= =- -3 2 1 ( 1)( 2)( 97)= = - - - -97( 1) 3!97!= = - -6(97)!= = - -1sin( )arcsin,(0)4.1

11、sinxf xxfx= =+ + - - 求求0( )(0)(0)limxf xffx- -= =解解01sinarcsin1sinlimxxxxx- -+ += =1= =(2).先化簡(jiǎn)再求導(dǎo)先化簡(jiǎn)再求導(dǎo) 21sin,sin24392.6xyyPx+ + = =例例指指求求導(dǎo)導(dǎo)例例222sincos2sincosxxyxx+ += = 解解1tancot2xx= =+ +221seccsc2yxx= =- - ( )55,141( ),( )1nPxf xfxx- -= = + + 求求例5指導(dǎo)例5指導(dǎo)2(1)2( )1,11xf xxx-+-+=-=-+解解22( ),(1)f xx- -

12、= =+ +23( 1) 2 2!( )(1)fxx-= =+ +( )1( 1) 2!,( )(1)nnnnfxx+ +- - = =+ +(3).將乘除變加減再求導(dǎo)將乘除變加減再求導(dǎo)()(1,6)xxxyyx x-= =求求例例2xxxx xyx x- - - -= =解解111xxx= =- - - -23111122yxxx= = - - -+ +23ln,1xyyx= =+ +例例求求7 721lnln(1),32yxx= =- -+ +解解2132(1)yxx=-=-+ +(4). 冪指函數(shù)的導(dǎo)數(shù)冪指函數(shù)的導(dǎo)數(shù)22128,xxxyxdy= = =+ +求求例例22 ln21xxxy

13、e=+=+解解221(2 ln2 ln2)2ln2 2xxxxyxxxx= = + + + + 11(24ln2)xxdyydxdx=+=+22128,xxxyxdy= = =+ +求求例例212,xyx= =解解 令令取取對(duì)對(duì)數(shù)數(shù)1ln2 ln ,xyxx= =兩兩邊邊對(duì)對(duì) 求求導(dǎo)導(dǎo): :11112 ln2 ln2xxyxyx= = + + 211(2 ln2 ln2)xxxyxxx= = + + (5).隱函數(shù)的導(dǎo)數(shù)隱函數(shù)的導(dǎo)數(shù)9(157,(0),9)yexyyPe+ += =求求例例 總總習(xí)習(xí)題題二二(0)1,:0yyxeyyxy= = + + += =方方程程兩兩邊邊對(duì)對(duì) 求求導(dǎo)導(dǎo)解解

14、1(0)1:(0)yye= = = - -把把代代入入知知()(1)y 不不必必解解出出(1)式兩邊再對(duì)式兩邊再對(duì)x求導(dǎo)：求導(dǎo)： 2( )0(,)2)yyeye yyyxyy+ + + + += = 不不必必化化簡(jiǎn)簡(jiǎn) 不不必必解解出出1(0)1,(0)(2):yye= = = - -把把代代入入式式得得211()(0)2()0eeyee-+-=-+-=222211(0)yeee= =- -= =(6).參數(shù)方程所確定的函數(shù)的導(dǎo)數(shù)參數(shù)方程所確定的函數(shù)的導(dǎo)數(shù)2222210(, 157,10.(2)ln 1,arctand yd xxtytdxPdy =+=+ = = 例總習(xí)題例總習(xí)題求求二二221

15、112121dtdytdxttdtt+ +=+ +解解22()dydd ydxdxdx= =2211dtttdtt- -= =+ +231tt+ += -= -1dxtdydydx=22()dxddyd xdydy= =211dtdtt= =+ +21t=+=+ 2arcta11(222,4(2)n( ),.25tdyxtyg xytyedxCH= =- -= =- -+ += =設(shè)設(shè)由由確確定定 求求例例習(xí)習(xí)解解補(bǔ)補(bǔ)充充練練習(xí)習(xí)題題225tytyet- -+ += =兩兩邊邊對(duì)對(duì)解解求求導(dǎo)導(dǎo): :2220tdydyytyedtdt- +=- +=222tdyyedtty- -= =- -tt

16、ydydxx= =222211tyetyt- - -= =+ +22()(1)2(1)tyetty- -+ += =- -(7).籠統(tǒng)函數(shù)的高階導(dǎo)數(shù)籠統(tǒng)函數(shù)的高階導(dǎo)數(shù) 設(shè)設(shè)f f( (x x) )可可導(dǎo)導(dǎo), ,y y= =f f( (l ln nx x) )例例, ,求求1 12 2y y1(ln )yfxx = = 解解2211(ln )(ln ) ()yfxfxxx=+ -=+ -21(ln )(ln )fxfxx = =- -213( ),(),x xy ey- - = =例例設(shè)設(shè)可可導(dǎo)導(dǎo)求求2 ()() ()xxxy e ee- - - - = = - -解解222222xxxy e

17、ee- - - - = =+ + +3.導(dǎo)數(shù)的應(yīng)用導(dǎo)數(shù)的應(yīng)用1中值定理證等式、證不等式、證方程根的存在性中值定理證等式、證不等式、證方程根的存在性0, ( ) , , ),:( , ),( )( )(14(224,6)lnab f xa ba ba bbf bf aPf a -=-=設(shè)在上設(shè)在上連續(xù) 在(連續(xù) 在(例總習(xí)題三例總習(xí)題三內(nèi)可導(dǎo) 證明 存在使內(nèi)可導(dǎo) 證明 存在使證證1倒推法要證等式倒推法要證等式( )( )( )lnln1f bf af ba - -= =- -變形為變形為( )ln , , F xxxa b= = 令令1( )0,( , )F xxa bx = = 對(duì)對(duì)f(x)、

18、F(x) 應(yīng)用應(yīng)用Cauchy定理，定理，( )( )( )( , ),( )( )( )f bf af a bF bF aF - - = = - -使使得證得證.證證2原函數(shù)法要證等式化為原函數(shù)法要證等式化為 ( )( )( )ln 0 x bf bf axf xa= = - - -= =1 ( )( )( )ln 0 x bf bf af xxa= = - - -= =或或 ( )( )ln( )ln 0 x bf bf axf xa= = - - -= =即即( ) ( )( )ln( )lnb xf bf axf xa= =- - -令令輔輔助助函函數(shù)數(shù)( )( )ln( )ln( )

19、 af baf ab b= =- -= =則則( ) , (0,0). xa bRollea 對(duì)對(duì)在在上上應(yīng)應(yīng)用用定定理理并并注注意意即即證證畢畢.( ) , ,( , ),( )( ),:( , ),(05)1f xa ba bf af ba bf = = 設(shè)不恒為常數(shù)的函數(shù)在上連續(xù) 在設(shè)不恒為常數(shù)的函數(shù)在上連續(xù) 在內(nèi)可導(dǎo) 且證明 在內(nèi)至少存在內(nèi)可導(dǎo) 且證明 在內(nèi)至少存在一點(diǎn)使得一點(diǎn)使得例例( )( )( )( , )f bf af xa b= =因因且且在在上上不不證證恒恒為為常常數(shù)數(shù)( , ),( )( )ca bf cf a 故故使使( )( ), , f cf aa clagrang

20、e 若若在在上上應(yīng)應(yīng)用用定定理理，( )( )( , )( )0f cf aa cf ca- - = = - -則則使使( )( ), , f cf ac blagrange 若若在在上上應(yīng)應(yīng)用用定定理理，( )( )( , )( )0f bf cc bf bc- - = = - -則則使使abcxyo163113 21xxeex- - 例例 （習(xí)習(xí)題題， （）證證明明：當(dāng)當(dāng)時(shí)時(shí), ,( ),1,1xf xexLagrange= =令令上上證證在在應(yīng)應(yīng)用用定定理理(1)(1)(1)xe xeeeexxex - -= =- - -= =- - xeex( )ln ,1, f xxxLagrang

21、e= =令在上應(yīng)用令在上應(yīng)用證2證2定理定理1lnln1(1)1(1()1)xxxx-=-=-1ln1xxxxe- -由得由得,xxexeexe即即( )3xf xeex=-=-令令證證( )0,1xf xeex = =- - 當(dāng)當(dāng)時(shí)時(shí)1, ( )1, ( )(1)0 xf xxf xf = =時(shí)時(shí)單單調(diào)調(diào)增增加加時(shí)時(shí)0,xeex-即即xeex 故故2單調(diào)性、凹凸性單調(diào)性、凹凸性,:17(66,3)babPaeab設(shè)證明設(shè)證明例自測(cè)例自測(cè)lnln:lnln ,abbaabab 分分析析 取取對(duì)對(duì)數(shù)數(shù)ln( )( , 1)xF xF xxa bx= = 令令證證證證21ln( )xF xx- -

22、 = =( )0eaxbF x ( ) , F xa b在在( )( )abF aF b又又lnlnabab 即即lnlnbaab baab: lnlnbaab 分分析析 只只須須征征( )lnln()2f xxaaxxa=-=-證證 令令1( )l0()naxaeafaxxx -=-=-( )f xxa 在在時(shí)時(shí)單單調(diào)調(diào)增增加加( )( )0baf bf a = =時(shí)時(shí)有有l(wèi)nln0baab- - 即即lnlnbaab baab 1:0,arctan.218xxx + + 例例證證明明時(shí)時(shí)1( )arctan2f xxx= =+ +- -設(shè)設(shè)證證2222111( )01(1)f xxxxx

23、= = - -+ += = - - + + +0( )xf x時(shí)時(shí)1( )()lim(arctan)02xf xfxx+ + + + = =+ +- -= =1arctan2xx+ + 22:0,(191)ln(1.)xxxx-例例證明時(shí)證明時(shí)22( )(1)ln(1) ,(11)0 xxxx= =- - - -= =證證 令令易易知知211( )2 ln(1)2(1)2 ln2 xxxxxxxxxx = =+ +- - - -= =- - -+ +(1)0,1x = = = 是是駐駐點(diǎn)點(diǎn)22111( )2ln212ln1 xxxxxxx = =+ +- - + += =+ + +(1)20,

24、( )1,1( )(0,) xxx x = = = = =+ + 在在取取極極小小值值且且為為在在內(nèi)內(nèi)唯唯一一極極值值點(diǎn)點(diǎn)( )1(0,),(1)0 xx= =+ + = =在在處處取取得得內(nèi)內(nèi)的的最最小小值值 又又0, ( )(1)0 x x = =時(shí)時(shí)22(1)ln(1)xxx- - - -即即22:0,(191)ln(1.)xxxx-例例證明時(shí)證明時(shí)12( )lnln1121x xxxxx- -=-=-+=-=-+證證 令令222112( )0,0(1)(1)x xxxxx x+ + = =- -= = + + +當(dāng)當(dāng)0, ( )(1)0 x x = =時(shí)時(shí)又又01 , ( )0,1,

25、( )0 x xx x + + 當(dāng)當(dāng)時(shí)時(shí)當(dāng)當(dāng)時(shí)時(shí)2220,(1) ( )(1)ln(1)0 xx xxxx - -= =- - - - 時(shí)時(shí)22(1)ln(1)xxx- - - -即即:( )(1)ln(1) xxxx=+-=+-注 也可令注 也可令2120.xxyxedx- - -= = 求求例例的的極極值值21,0 xyxex- - = = =駐駐點(diǎn)點(diǎn)為為解解0,0;0,0.xyxy 時(shí)時(shí)時(shí)時(shí)0 x= = 為為極極小小值值點(diǎn)點(diǎn), ,極極小小值值為為: :220011111(0)(1)22xxyxedxee- - - - - -= = = - -= =- - 222222,0 xxxxyxexyex eex- - = =駐點(diǎn)為駐點(diǎn)為-2=(1-2),y (-2=(1-2),y (解2解20)=100)=100 x= = 為為極極小小值值點(diǎn)點(diǎn), ,極極小小值值為為: :