清華微積分高等數(shù)學(xué)第七講數(shù)與微分三

1、2021-11-71 作業(yè)作業(yè)p68 習(xí)題習(xí)題3.2 13(5). 14(6). 15(2). 16(3). p73 習(xí)題習(xí)題3.3 4(1). 5(5). 7(4). 9.p78 習(xí)題習(xí)題: 5(5) (6) (8). 6(3). 7. 8.復(fù)習(xí)復(fù)習(xí)p60p782021-11-72二、高階導(dǎo)數(shù)二、高階導(dǎo)數(shù)第七講第七講 導(dǎo)數(shù)與微分導(dǎo)數(shù)與微分( (三三) )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 （續(xù)）（續(xù)）2021-11-73一、導(dǎo)數(shù)與微分運(yùn)算法則一、導(dǎo)數(shù)與微分運(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2. 復(fù)合函數(shù)求導(dǎo)法則復(fù)合函數(shù)求導(dǎo)法則3. 反函數(shù)求導(dǎo)法反函數(shù)求導(dǎo)法4. 隱函

2、數(shù)求導(dǎo)法隱函數(shù)求導(dǎo)法5. 參數(shù)方程求導(dǎo)法參數(shù)方程求導(dǎo)法6. 對(duì)數(shù)微分法對(duì)數(shù)微分法2021-11-745. 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：例例aa 22021-11-752, 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：2021-11-760120(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

3、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 2021-11-77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 2021-11-782, 0sincos9 ttbytax求求橢橢圓圓：例例,24cos,4aaxt 時(shí)時(shí)當(dāng)當(dāng).4處處的的切切線線方方程程在在 t24sinbby )2,2(:0aam切切點(diǎn)點(diǎn)解解4tan: tdxdyk切切線線斜斜率率2021-11-79tabtatbtt

4、dxdycotsincos)()( ababdxdykt 44sincos4 :切切線線方方程程)2(2axabby bxaby2 即即2021-11-710的的導(dǎo)導(dǎo)數(shù)數(shù)求求冪冪指指函函數(shù)數(shù))()()(xvxuxf )(ln)()(xuxvexf 再應(yīng)用復(fù)合函數(shù)微分法（鏈?zhǔn)椒▌t）再應(yīng)用復(fù)合函數(shù)微分法（鏈?zhǔn)椒▌t）方法二方法二: 利用對(duì)數(shù)微分法利用對(duì)數(shù)微分法方法一方法一: )()( )(ln xfxfxf 6. 對(duì)數(shù)微分法對(duì)數(shù)微分法 )()ln()( xfxfxf2021-11-711yxyx 的的導(dǎo)導(dǎo)數(shù)數(shù)求求冪冪指指函函數(shù)數(shù)例例cos)(sin1得得到到求求導(dǎo)導(dǎo)兩兩邊邊對(duì)對(duì),xxxxxxyysi

5、ncoscos)ln(sin)sin(1 解解 兩邊取對(duì)數(shù)兩邊取對(duì)數(shù), 得得得得解解出出,y )ln(sinsinsincos)(sin2cosxxxxxyx 對(duì)數(shù)微分法對(duì)數(shù)微分法)ln(sincoslnxxy 2021-11-712yxxxxy 求求設(shè)設(shè)例例,)4)(3()2)(1(23)4ln() 3ln() 2ln() 1ln(31ln xxxxy4131211131 xxxxyy解解)41312111() 4)(3() 2)(1(31)(ln3 xxxxxxxxyyy2021-11-713例例3解解 求切線斜率求切線斜率(x,y)處切線方程：處切線方程：。間間的的線線段段長(zhǎng)長(zhǎng)度度等等于

6、于常常數(shù)數(shù)兩兩坐坐標(biāo)標(biāo)軸軸之之上上任任一一點(diǎn)點(diǎn)處處的的切切線線介介于于證證明明星星形形線線323232ayx 032323131 yyx33xyy )(33xxxyyy 2021-11-714化為截距式化為截距式線段長(zhǎng)度：線段長(zhǎng)度：常數(shù)常數(shù)3232323333)(axyyxxyyxxy 13232 yayxaxaayaxal 2232232)()(2021-11-715近近似似計(jì)計(jì)算算微微分分的的簡(jiǎn)簡(jiǎn)單單應(yīng)應(yīng)用用xxfxfxxfdyyx )()()(,1000即即有有時(shí)時(shí)當(dāng)當(dāng))()()()()()()(000000 xxxfxfxfxxfxfxxf 或或xffxfx )0()0()(,00有有

7、時(shí)時(shí)當(dāng)當(dāng)2021-11-716例例的的近近似似值值求求2160cos )18060123cos(2160cos 3,cos)(0 xxxf令令)10800123cos( 解解10800120 xxx2021-11-717則則有有3sin)(,3cos)(00 xfxf得得由由近近似似公公式式)()()()(000 xxxfxfxf 4970. 01080012232110800123sin3cos2160cos 2021-11-718xxxxxarctgxxxxxxxxxxexffxfxx2111,1)1(,arcsintan,sin)1ln(,1)0()0()(,1 可得下列近似公式：可得下

8、列近似公式：利用利用時(shí)時(shí)當(dāng)當(dāng)2021-11-71922),(),(,)(,)()(dxydxyxfxxfxxfxf或或或或記記作作的的二二階階導(dǎo)導(dǎo)數(shù)數(shù)在在稱稱為為函函數(shù)數(shù)的的導(dǎo)導(dǎo)數(shù)數(shù)在在的的導(dǎo)導(dǎo)函函數(shù)數(shù)函函數(shù)數(shù) 二、高階導(dǎo)數(shù)二、高階導(dǎo)數(shù)xxfxxfxfx )()(lim)(0 即即（一）高階導(dǎo)數(shù)定義（一）高階導(dǎo)數(shù)定義2021-11-72033),(),(,)(,)()(dxydxyxfxxfxxfxf或或或或記記作作的的三三階階導(dǎo)導(dǎo)數(shù)數(shù)在在稱稱為為函函數(shù)數(shù)導(dǎo)導(dǎo)數(shù)數(shù)的的在在的的二二階階導(dǎo)導(dǎo)函函數(shù)數(shù)函函數(shù)數(shù) xxfxxfxfnnxn )()(lim)()1()1(0)( 即即nnnndxydxyx

9、fnxxfxnxf或或或或記記作作階階導(dǎo)導(dǎo)數(shù)數(shù)的的在在稱稱為為函函數(shù)數(shù)的的導(dǎo)導(dǎo)數(shù)數(shù)階階導(dǎo)導(dǎo)函函數(shù)數(shù)在在的的函函數(shù)數(shù)),(),(,)(,)1()()()( 2021-11-721)(tss 變變速速直直線線運(yùn)運(yùn)動(dòng)動(dòng)：瞬瞬時(shí)時(shí)速速度度一一階階導(dǎo)導(dǎo)數(shù)數(shù)：)()(tvts 瞬瞬時(shí)時(shí)加加速速度度二二階階導(dǎo)導(dǎo)數(shù)數(shù)：)()(tats 二階導(dǎo)數(shù)的物理意義二階導(dǎo)數(shù)的物理意義2021-11-722)(), 2, 1(1nnynxy求求例例 1 nnxy2)1( nxnny!)(nyn 解解用數(shù)學(xué)歸納法可以證明用數(shù)學(xué)歸納法可以證明2021-11-723aayxln )(),1, 0(2nxyaaay求求例例 2)(

10、lnaayx nxnaay)(ln)( xnxee )()(特特例例：用數(shù)學(xué)歸納法可以證明用數(shù)學(xué)歸納法可以證明解解2021-11-724.sin3階階導(dǎo)導(dǎo)數(shù)數(shù)的的求求例例nx)2sin()(sin)( nxxn)2sin(cos)(sin xxxxxx )2sin()(sin )22sin()2cos( xx解解xxx )22sin()(sin )23sin()22cos( xx用數(shù)學(xué)歸納法用數(shù)學(xué)歸納法2021-11-725).0(, 3/)0(,01cos22)(4yyyexfyx 求求且且確確定定由由方方程程函函數(shù)數(shù)例例 求求導(dǎo)導(dǎo)方方程程兩兩邊邊對(duì)對(duì) x) 1 (0sin22 yyex得得

11、代代入入將將),1(, 3/)0(, 0 yx32)0( y解解得得求求導(dǎo)導(dǎo)式式兩兩邊邊再再對(duì)對(duì),)1(x得得代代入入將將),2(,3/2)0(, 3/)0(, 0 yyx 9310)0( y)2(0sincos2 yyyyex2021-11-726).(,cossincosln)(5xytttytxxfy 求求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)例例tttttttttxtyxycoscos/sinsincoscos)()()( 解解xxydxdxy)()( ttxytxxycos)(cosln)(由由參參數(shù)數(shù)方方程程確確定定2021-11-727xxyxy )()( ttttt)cos(

12、ln)cos( tttxy)cos()(: 注注意意tttttcos/sinsincos tttttsincossincos2 )( )(txxyt 2021-11-728則則階階導(dǎo)導(dǎo)數(shù)數(shù)有有設(shè)設(shè)函函數(shù)數(shù),)(),(nxvxu)()()()()1(nnnvuvu )()()()2(nnucuc )()(0)()()3(kknnkknnvucvu ),()0()0(vvuu 其其中中式式稱稱為為萊萊布布尼尼茲茲公公式式)3(（二）高階導(dǎo)數(shù)性質(zhì)（二）高階導(dǎo)數(shù)性質(zhì)2021-11-729)(32,15nxyexy求求設(shè)設(shè)例例 則則令令,23xveux , 2,2 vxv由由萊萊布布尼尼茲茲公公式式得得

13、), 2, 1(33)(nkeuxkk 0)()4( nvvv)()(! 2) 1()()()()(2)2(32)1(32)(3)(23)( xennxenxexeynxnxnxnxn)1(693232 nnnxxexn解解2021-11-730vuvuvu )(xuxuyy 乘乘、除除四四則則計(jì)計(jì)算算法法則則特特別別注注意意)1(復(fù)復(fù)合合求求導(dǎo)導(dǎo)法法則則)2(2)(vvuvuvu 函函數(shù)數(shù)關(guān)關(guān)系系注注意意分分析析清清楚楚小結(jié)小結(jié)1 導(dǎo)數(shù)計(jì)算導(dǎo)數(shù)計(jì)算要要求求反反函函數(shù)數(shù)求求導(dǎo)導(dǎo)公公式式)( 1)()3(1xfyf 0)( xf2021-11-731)( )() )()(txxyxyxytx 求求二二階階導(dǎo)導(dǎo)數(shù)數(shù)注