清華大學(xué)微積分高等數(shù)學(xué)課件第6講導(dǎo)數(shù)與微分二

1、2021-12-301 作作 業(yè)業(yè) 6(3) (6) (9) (11) (14) (17). 9(4) (8) (15) (21). 10(8). 11(2). 12(2). p67 習(xí)題習(xí)題3.2 2021-12-302二、高階導(dǎo)數(shù)二、高階導(dǎo)數(shù)第六講第六講 導(dǎo)數(shù)與微分導(dǎo)數(shù)與微分( (二二) )一、導(dǎo)數(shù)與微分的運(yùn)算法則一、導(dǎo)數(shù)與微分的運(yùn)算法則2021-12-303一、導(dǎo)數(shù)與微分的運(yùn)算法則一、導(dǎo)數(shù)與微分的運(yùn)算法則1. 四則運(yùn)算求導(dǎo)法則四則運(yùn)算求導(dǎo)法則則則可可導(dǎo)導(dǎo)在在設(shè)設(shè)函函數(shù)數(shù),)(),(xxvxu且且可可導(dǎo)導(dǎo)在在函函數(shù)數(shù),)()()1(xxvxu )()( )()(xvxuxvxu 且且為為常

2、常數(shù)數(shù)可可導(dǎo)導(dǎo)在在函函數(shù)數(shù)),()()2(cxxuc)( )(xucxuc 2021-12-304且且可可導(dǎo)導(dǎo)在在函函數(shù)數(shù),)()()3(xxvxu )()()()( )()(xvxuxvxuxvxu 且且可可導(dǎo)導(dǎo)在在函函數(shù)數(shù),)()()4(xxvxu2)()()()()()()(xvxvxuxvxuxvxu )0)( xv2021-12-305)()(xvxuy 設(shè)設(shè))()()()(xxvxuxxvxxu vxuxxvu )()()()()()(xvxuxxvxu 證證 (3)xvxuxxvxuxy )()()()()()()()(limlim00 xvxuxvxuxvxuxxvxuxyyx

3、x 可導(dǎo)必連續(xù)可導(dǎo)必連續(xù))()()()(xvxuxxvxxuy 2021-12-306的的導(dǎo)導(dǎo)數(shù)數(shù)求求函函數(shù)數(shù)例例2sinlncos24735 xxxxyxxxx1sin212524 解解)2(sin)(ln)(cos2)( 4)(35 xxxx)2sinlncos24(35 xxxxy2021-12-307)cossin()(tan xxxxxxxx2cos)(cossincos)(sin .seccos1cos)sin(sincoscos222xxxxxxx 的導(dǎo)數(shù)的導(dǎo)數(shù)求函數(shù)求函數(shù)例例xxftan)(8 解解xxx22cos1sec)(tan 2021-12-3082、復(fù)合函數(shù)導(dǎo)數(shù)公式、

4、復(fù)合函數(shù)導(dǎo)數(shù)公式（1）復(fù)合函數(shù)微分法（鏈?zhǔn)椒▌t）復(fù)合函數(shù)微分法（鏈?zhǔn)椒▌t）且且也也可可導(dǎo)導(dǎo)在在點(diǎn)點(diǎn)則則復(fù)復(fù)合合函函數(shù)數(shù)可可導(dǎo)導(dǎo)在在點(diǎn)點(diǎn)函函數(shù)數(shù)可可導(dǎo)導(dǎo)在在點(diǎn)點(diǎn)設(shè)設(shè)函函數(shù)數(shù),)(,)(,)(xxgfyxxguuufy dxdududydxdy 或或)()()(xgxgfdxxgdf 2021-12-309 證證 xyx 0lim xuuyxuuyxux 000limlimlimdxdududy )()(xgxgf 0,0 xx 時(shí)時(shí)當(dāng)當(dāng)不能保證中間變量的增量不能保證中間變量的增量)()(xgxxgu 總不等于零總不等于零上面的證法有沒有問題？上面的證法有沒有問題？2021-12-3010 證證

5、可可導(dǎo)導(dǎo))(ufy )(ufuy)0lim(0 u)(lim0ufuyu 上上式式化化為為時(shí)時(shí)當(dāng)當(dāng),0 u )1()(uuufy 0)()(,0 ufuufyu 時(shí)時(shí)當(dāng)當(dāng)(1) 式仍然成立！式仍然成立！xuxuufxy )(2021-12-3011xuxuufxyxxxx 0000limlimlim)(lim )()(lim)(0 xgufxydxxgdfx 0limlim00 ux)()()(xgufdxxgfd 連連續(xù)續(xù)可可導(dǎo)導(dǎo))()(xguxgu 00ux 2021-12-3012（2）微分的形式不變性）微分的形式不變性(復(fù)合函數(shù)微分法則復(fù)合函數(shù)微分法則)且且其其微微分分為為也也可可微微

6、則則復(fù)復(fù)合合函函數(shù)數(shù)函函數(shù)數(shù)均均為為可可微微和和設(shè)設(shè)函函數(shù)數(shù),)(,)()(xgfyxguufy dxxuufduufdy)( )( )( xxgfdyx )(證證xxgxgf )()( duuf )(有有時(shí)時(shí)當(dāng)當(dāng),)(xxgu xxxgdu )(xdx 2021-12-3013我我們們將將微微分分寫寫成成因因此此對(duì)對(duì)于于自自變變量量,xdxxfxxfxdf)()()( dxxfxdf)()( uduxxgu ,)(時(shí)時(shí)當(dāng)當(dāng)不不能能將將微微分分寫寫成成對(duì)對(duì)于于中中間間變變量量),(xuu uufxudf )()( dxxuufduufxudf)( )( )()( 的的函函數(shù)數(shù)，微微分分形形式式

7、不不變變還還是是中中間間變變量量是是自自變變量量不不論論uxy但有但有 微分的微分的形式不變性形式不變性2021-12-3014.11123的的導(dǎo)導(dǎo)數(shù)數(shù)求求函函數(shù)數(shù)例例 xxy解解 11112321xxdxdxxdxdy221)1(21123 xxx2521)1()1(3 xx2021-12-301542,ln xvtgvuuy設(shè)設(shè)xvuxxtgvuy)42()()(ln .)42lntan(2的的導(dǎo)導(dǎo)數(shù)數(shù)求求函函數(shù)數(shù)例例 xy解解21cos112 vu21)(cos1)(142242 xxtg)sin(12 xxcos1 xsec 2021-12-3016.)1ln(32的的導(dǎo)導(dǎo)數(shù)數(shù)求求函函

8、數(shù)數(shù)例例 xxyxxxxxxy)1(1122 1111112222 xxxxxx)1(11122xxxx 解解)1(121111222xxxxx 2021-12-3017.ln4的的導(dǎo)導(dǎo)數(shù)數(shù)求求函函數(shù)數(shù)例例xy 時(shí)時(shí)當(dāng)當(dāng)時(shí)時(shí)當(dāng)當(dāng)0, )ln(0,lnlnxxxxxyxxyx1)(ln,0 時(shí)時(shí)當(dāng)當(dāng) )ln(,0 xyx時(shí)時(shí)當(dāng)當(dāng))0(1)(ln xxxxxx1)(1 解解.lnln同同的的導(dǎo)導(dǎo)數(shù)數(shù)公公式式有有相相與與xx)()( )(ln)(lnxfxfxfxf 2021-12-3018.,1中中間間變變量量都都便便于于求求導(dǎo)導(dǎo)應(yīng)應(yīng)使使每每一一個(gè)個(gè)地地選選取取中中間間變變量量恰恰當(dāng)當(dāng)在在于于分分

9、析析清清楚楚函函數(shù)數(shù)關(guān)關(guān)系系關(guān)關(guān)鍵鍵：復(fù)復(fù)合合函函數(shù)數(shù)求求導(dǎo)導(dǎo)數(shù)數(shù)時(shí)時(shí)注注意意.,2就就用用什什麼麼求求導(dǎo)導(dǎo)法法則則什什麼麼運(yùn)運(yùn)算算碰碰到到四四則則運(yùn)運(yùn)算算的的函函數(shù)數(shù)關(guān)關(guān)系系時(shí)時(shí)又又有有：當(dāng)當(dāng)遇遇到到既既有有復(fù)復(fù)合合運(yùn)運(yùn)算算注注意意2021-12-30193. 反函數(shù)求導(dǎo)法則反函數(shù)求導(dǎo)法則)(1 )(,)()(, 0)(,11xfyfxfyyyfxxfxfx 且且可可導(dǎo)導(dǎo)在在反反函函數(shù)數(shù)則則它它的的且且可可導(dǎo)導(dǎo)在在單單調(diào)調(diào)且且嚴(yán)嚴(yán)格格的的某某鄰鄰域域連連續(xù)續(xù)設(shè)設(shè)函函數(shù)數(shù)在在2021-12-3020的導(dǎo)數(shù)的導(dǎo)數(shù)求函數(shù)求函數(shù)例例xxfyarcsin)( 解解2211 yxyxxysin,)1,

10、1(arcsin 存存在在反反函函數(shù)數(shù)增增加加且且嚴(yán)嚴(yán)格格上上連連續(xù)續(xù)在在yyxcos1)(sin1)(arcsin 2211sin11xy 由反函數(shù)由反函數(shù)求導(dǎo)法則求導(dǎo)法則2021-12-30214. 隱函數(shù)求導(dǎo)法隱函數(shù)求導(dǎo)法定義：（隱函數(shù)）定義：（隱函數(shù)）.0),()(,0),(,.,的的隱隱函函數(shù)數(shù)確確定定是是方方程程或或關(guān)關(guān)系系則則稱稱此此對(duì)對(duì)應(yīng)應(yīng)對(duì)對(duì)應(yīng)應(yīng)唯唯一一的的由由方方程程若若設(shè)設(shè)有有非非空空數(shù)數(shù)集集 yxfxfyfyyyxfxxryx0)(, xfxfxx有有的的解解必必是是方方程程確確定定的的隱隱函函數(shù)數(shù)由由方方程程注注意意0),()(0),( yxfxfyyxf2021-1

11、2-3022.),(0),(可可導(dǎo)導(dǎo)并并且且函函數(shù)數(shù)隱隱函函數(shù)數(shù)能能夠夠確確定定假假定定方方程程fxfyyxf ?,)(如如何何求求出出導(dǎo)導(dǎo)數(shù)數(shù)的的情情況況下下問問題題：在在不不解解出出顯顯式式xfy 隱函數(shù)求導(dǎo)問題的提法隱函數(shù)求導(dǎo)問題的提法2021-12-3023.,0)(,(),(,0),(xyxxyxfxxfyxyyxf 解解出出求求導(dǎo)導(dǎo)兩兩邊邊對(duì)對(duì)的的恒恒等等式式：關(guān)關(guān)于于于于是是方方程程可可看看成成的的函函數(shù)數(shù)：看看成成把把中中在在方方程程.,求求導(dǎo)導(dǎo)法法則則因因此此需需要要應(yīng)應(yīng)用用復(fù)復(fù)合合函函數(shù)數(shù)的的函函數(shù)數(shù)是是要要注注意意注注意意：左左端端求求導(dǎo)導(dǎo)時(shí)時(shí)xy 隱函數(shù)求導(dǎo)法隱函數(shù)求導(dǎo)

12、法2021-12-3024得得求求導(dǎo)導(dǎo)方方程程兩兩邊邊對(duì)對(duì),x)2(02sincos3cos)(22223 xxxxxyxyyyexy.),(01cos 123xxyyxfyxxye 求求隱隱函函數(shù)數(shù)確確定定由由方方程程例例解解)1(0)1()cos()(23 xxeyyexyxy得得解解出出,y )1(sincos6cos2222223xyeeyxxxxyxyxy ?)0(: y問問0)0(1)0( yy2021-12-30255. 參數(shù)方程求導(dǎo)法參數(shù)方程求導(dǎo)法參參數(shù)數(shù)方方程程)1(2, 0sincos1 ttbytax橢橢圓圓：例例0,)cos1()sin(2 atayttax擺擺線線：例

13、例aa 22021-12-30262, 0sincos333 ttaytax星星形形線線：例例a內(nèi)旋輪線內(nèi)旋輪線0,323232 aayx隱隱函函數(shù)數(shù)方方程程：2021-12-30270120(2) 參數(shù)方程求導(dǎo)法參數(shù)方程求導(dǎo)法?dxdy如如何何求求).()(, 0)(,)(),(1xttxttt 存存在在可可導(dǎo)導(dǎo)的的反反函函數(shù)數(shù)且且都都存存在在設(shè)設(shè)確確定定由由參參數(shù)數(shù)方方程程：設(shè)設(shè)函函數(shù)數(shù) )()()(tytxxfy 2021-12-3028的的復(fù)復(fù)合合函函數(shù)數(shù)成成為為通通過過xty)(ty 分析函數(shù)關(guān)系分析函數(shù)關(guān)系:)()(1xttx )(1xy 利用復(fù)合函數(shù)和反函數(shù)微分法利用復(fù)合函數(shù)和反函數(shù)微分法, 得得)()(ttdtdxdtdydxdtdtdydxdy 20