高等數(shù)學(xué)微積分第3章第3節(jié)反復(fù)隱的導(dǎo)數(shù)ppt課件

1、第三節(jié)第三節(jié) 反函數(shù)、復(fù)合函數(shù)、隱函數(shù)的導(dǎo)數(shù)反函數(shù)、復(fù)合函數(shù)、隱函數(shù)的導(dǎo)數(shù)一一.反函數(shù)的導(dǎo)數(shù)反函數(shù)的導(dǎo)數(shù)設(shè)函數(shù)設(shè)函數(shù) y=f (x) 在點(diǎn)在點(diǎn)x處有不等于零處有不等于零的導(dǎo)數(shù)的導(dǎo)數(shù)),(xf 并且其反函數(shù)并且其反函數(shù))(1yfx 在相應(yīng)在相應(yīng)點(diǎn)處連續(xù)點(diǎn)處連續(xù),那么那么 )(1 yf存在存在,并且并且)(1 )(1xfyf 或或. )(1)(1 yfxf)(1 )(1xfyf yxy0lim xyy1lim0 xyx0lim1)(1xf 從而從而.)(1 )(1xfyf 因因 yxxy 1故故)0( y xyy0lim1設(shè)設(shè))(1yfx 證證例例1 求指數(shù)函數(shù)求指數(shù)函數(shù))1, 0( aaayx的

2、導(dǎo)數(shù)的導(dǎo)數(shù).解解xay 的反函數(shù)為的反函數(shù)為yxalog )(xa)(log1 yaeyalog11 ayln aaxln 特別地特別地.)(xxee 例例2 求求)11(arcsin xxy的導(dǎo)數(shù)的導(dǎo)數(shù).解解的反函數(shù)為的反函數(shù)為yxsin )(arcsin x)(sin1 yycos1 y2sin11 .112x xyarcsin )(arccosx211)(arcsinxx )(arctanx )cot(xarc211x 211x 211x 例例3 求求xarcxeayxxcotarctan 的導(dǎo)數(shù)的導(dǎo)數(shù).解解)cotarctan( xarcxeayxx)cot()(arctan)( xa

3、rcxeaxx221111)()(xxaeeaxxxx xxxxaeeaa ln).ln1(aeaxx )(cos8(x )(sin7(x )(ln4(x )(log3(xa )(2( x c )1( )(5(xa )(6(xe )(tan9(x )(cot10(x )(sec11(x )(csc12(x )(arcsin13(x )(arccos14(x )(arctan15(x )cot)(16(xarc基本求導(dǎo)公式基本求導(dǎo)公式二二.復(fù)合函數(shù)的導(dǎo)數(shù)復(fù)合函數(shù)的導(dǎo)數(shù)假如假如)(xu 在點(diǎn)在點(diǎn)x處有導(dǎo)數(shù)處有導(dǎo)數(shù)),(xdxdu )(ufy 在對(duì)應(yīng)點(diǎn)在對(duì)應(yīng)點(diǎn)u處有導(dǎo)數(shù)處有導(dǎo)數(shù)),(ufdudy 則

4、復(fù)合函數(shù)則復(fù)合函數(shù))(xfy 在點(diǎn)在點(diǎn)x處的導(dǎo)數(shù)也存在處的導(dǎo)數(shù)也存在,而且而且dxdududydxdy ).()()(xufxf 或或dxdududydxdy xyxuuy xyx0limxuuyx 0limxuuyxx 00limlimxuuyxu 00limlimdxdududy .dxdududydxdy 故故)0( u證證例例4求求2)12( xy的導(dǎo)數(shù)的導(dǎo)數(shù).解解1442 xxy y. 48 x解解令令12 xu那那么么2uy dxdy)(2 u22 u).12(4 x1)4()4(2 xx)12( x解解 y)12(2 x)21( x).12(4 x例例5 求求nxycos 的導(dǎo)數(shù)

5、的導(dǎo)數(shù).解解令令nxu 那那么么uycos dxdy)(cos unu )sin(.sinnxn 解解 y)( nx)sin(nx )( nxnnx )sin(.sinnxn 例例6求求)ln(22axxy 的導(dǎo)數(shù)的導(dǎo)數(shù).解解 y221axx 221axx 221axx ./122ax 221axx )(22 axx)(1 22 ax 1 22211 ax 2221ax )(22 ax)02( x例例7求求)310lncos(2xy 的導(dǎo)數(shù)的導(dǎo)數(shù).解解 y)310cos(12x )310tan(2x ).310tan(62xx )310cos(12x )310cos(2 x)310sin(2x

6、 )310(2 x)310(2 x例例8求求)1, 0(log aaxya的導(dǎo)數(shù)的導(dǎo)數(shù).解解0 xxyalog exyalog1 0 x)(logxya yexalog1 故故.log1)(logexxaa exalog1 )( x三三.抽象復(fù)合函數(shù)的導(dǎo)數(shù)抽象復(fù)合函數(shù)的導(dǎo)數(shù) )(xf xux)(xf )(x 例例9知知)(uf可導(dǎo)可導(dǎo), 求求, )(ln xf ,)( naxf .)( naxf解解 )(lnxf)(ln xf )(ln)/1(xfx )(naxf)(naxf )()(1nnaxfaxn )(naxf naxf)(1)( naxfn1)( naxfn).()(1axfaxfnn

7、 )(ln x)( nax1)( naxn)( ax )( axf)(axf )( ax四四.分段函數(shù)的導(dǎo)數(shù)分段函數(shù)的導(dǎo)數(shù)例例10 已知函數(shù)已知函數(shù) xf213 x33 x0 x10 x1 x求求).(xf 解解因因)0(, 3)1(ff 不存在不存在故故 xf00 x310 x23x1 x分段函數(shù)求導(dǎo)函數(shù)時(shí)注意分段函數(shù)求導(dǎo)函數(shù)時(shí)注意(1)每一段內(nèi)求導(dǎo)用法則求每一段內(nèi)求導(dǎo)用法則求,(2)分界點(diǎn)求導(dǎo)用定義求分界點(diǎn)求導(dǎo)用定義求.五五.隱函數(shù)的導(dǎo)數(shù)隱函數(shù)的導(dǎo)數(shù)例例11 求由方程求由方程123 yx所確定的隱函數(shù)所確定的隱函數(shù)的導(dǎo)數(shù)的導(dǎo)數(shù).)(xfy 解解方程兩邊作為方程兩邊作為x的函數(shù)同時(shí)求導(dǎo)的函數(shù)

8、同時(shí)求導(dǎo)1)()(23 yx23x.232yxy 得得即即故故yy 20 2yyx xy )(2yy )(2 xy )(xfy 0),(yxF求求.y 例例12 求由方程求由方程0 xyeyx所確定的隱函數(shù)所確定的隱函數(shù)的導(dǎo)數(shù)的導(dǎo)數(shù).)(xfy 解解方程兩邊作為方程兩邊作為x的函數(shù)同時(shí)求導(dǎo)的函數(shù)同時(shí)求導(dǎo)得得0)()( xyeyx即即yxe 故故.xyxyyeyex (1)y )(yxy 0 例例13 求由方程求由方程yxyxarctanln22 所確定的所確定的的導(dǎo)數(shù)的導(dǎo)數(shù).)(xfy 解解 方程兩邊作為方程兩邊作為x的函數(shù)同時(shí)求導(dǎo)的函數(shù)同時(shí)求導(dǎo)得得)(arctan)ln(2122 yxyx即

9、即)(2122yx 故故.xyxyy 隱函數(shù)隱函數(shù))22(yyx 2)(11yx 2yxyy 例例14 求曲線求曲線422 yxyx上點(diǎn)上點(diǎn)(2,2)處的處的的切線方程的切線方程.解解 方程兩邊作為方程兩邊作為x的函數(shù)同時(shí)求導(dǎo)的函數(shù)同時(shí)求導(dǎo)得得02)(2 yyyxyx故故xyyxy 221)2,2( y所以切線方程為所以切線方程為)2( 12 xy即即. 04 yx例例15 設(shè)球半徑設(shè)球半徑R以以2厘米厘米/秒等速度增加秒等速度增加,求當(dāng)球半徑求當(dāng)球半徑R=10厘米時(shí)厘米時(shí),其體積其體積V增加的速度增加的速度.解解334RV dtdVdtdRRR )34(3 dtdRR 24 )34(3Rdt

10、d .800210 dtdRRdtdV例例16 設(shè)設(shè)322)3(31xxxxy ,求求.y 解解 將等式兩邊取對(duì)數(shù)將等式兩邊取對(duì)數(shù))3ln(2)3ln(31)1ln(ln2lnxxxxy 方程兩邊作為方程兩邊作為x的函數(shù)同時(shí)求導(dǎo)的函數(shù)同時(shí)求導(dǎo) yy199311122 xxxxyy.99311122 xxxx六六.對(duì)數(shù)求導(dǎo)法對(duì)數(shù)求導(dǎo)法x2x 1131 x 31()32x 例例16 設(shè)設(shè)xxycos)(cot ,求求.y 解解 將將xxycos)(cot 兩邊取對(duì)數(shù)兩邊取對(duì)數(shù)xxycotlncosln 方程兩邊作為方程兩邊作為x的函數(shù)同時(shí)求導(dǎo)的函數(shù)同時(shí)求導(dǎo) yy1 y.csccotlnsin)(c

11、otcosxxxxx )sin(x xcotln xcos )csc(2x xcot1 )csc(cotln)sin(xxxy )(ty )(tx )(1xy 求求? dxdy dxdydxdtdtdy dtdxdtdy1 )(1)(tt .)()(tt )(1xy 是由是由)(ty )(1xt 復(fù)合而成復(fù)合而成解解七七.由參數(shù)方程確定的函數(shù)的求導(dǎo)法由參數(shù)方程確定的函數(shù)的求導(dǎo)法例例19 橢圓的參數(shù)方程為橢圓的參數(shù)方程為tbysin taxcos 橢圓在橢圓在4 t的對(duì)應(yīng)點(diǎn)的對(duì)應(yīng)點(diǎn)M0處的切線處的切線方程和法線方程方程和法線方程.,求求解解4 t時(shí)時(shí), 橢圓上的對(duì)應(yīng)點(diǎn)橢圓上的對(duì)應(yīng)點(diǎn)),(000yxM為為24cos0aa