第二節(jié)求導(dǎo)法則與初等函數(shù)求導(dǎo)ppt課件

1、2( )( ) ( )( )( ).( )( )u xux v xu x vxv xvxvuvuvu )(, )()()(xvxuxfhxfhxfxfh)()(lim)(0hxvxuhxvhxuh)()()()(lim0)()()()(xvxuxvxuhhxuh )(lim0)(xu)(hxvhxv)( )(xu)(hxvwvuwvu)(sin(tan )cosxyxx解解2(sin ) cossin (cos )cosxxxxx22222cossin1sec.coscosxxxxx21(cos )(sec )coscosxyxxx解解2sinsec tan ,cosxxxx0)( y)(yx

2、)(1)(yxf1,arcsinxy ,sin yx , )2,2(y)(arcsinx)(sinyycos1y2sin11211x )(arccosx211x0cosyxyarcsinxyarctanyy2sec)(tan內(nèi)連續(xù)可導(dǎo)。，在22tanyx)0(sec2y)(arctanxytan1y2sec1y2tan11211x211)arccot(xxyxtanxyarctan )arcsin(x211x )arccos(x211x )arctan(x211x )cotarc(x211x)(xgu )(ufy )(xgu fy )(xg( )( ).dydydy duf ug xdxdxd

3、u dx或或.sinln的導(dǎo)數(shù)的導(dǎo)數(shù)求函數(shù)求函數(shù)xy .sin,lnxuuy dxdududydxdy xucos1 xxsincos xcot .)1(102的的導(dǎo)導(dǎo)數(shù)數(shù)求求函函數(shù)數(shù) xyxu2109.)1(2092 xx. 1,210 xuuydxdududydxdy )(, )(, )(xhvvguufyxydd)()()(xhvgufyuvxuyddvuddxvdd.1sin的導(dǎo)數(shù)的導(dǎo)數(shù)求函數(shù)求函數(shù)xey xvvueyu1,sin,.1cos11sin2xexx 21cosxvedxdvdvdududydxdyu1sin,.xyey求求11sinsin1()(sin)xxyeex解解1

4、sin11cos( )xexx1sin211cos. xexx)1()1(10292 xxdxdyxx2)1(1092 .)1(2092 xx.)1(102的的導(dǎo)導(dǎo)數(shù)數(shù)求求函函數(shù)數(shù) xy1cos()cos()xxyee解解1 sin() ()tan().cos()xxxxxeeeee 321 2,.yxy求求12222331(12) (12)(12)3解解 yxxx2234.3 (1 2)xx1 cosln,sinxyx.y求1ln 1 coslnsin 2yxx1sincos2 1cossinxxyxx 1 cos2sin (1 cos )xxx1csc2x2sin,yfx xuufy2si

5、n xufy2sin xxufcossin2 xuf2sinxxf2sinsin2)(uf.y1(11) (ln );xx 21(14) (arccos );1xx 1(12) (log);lnaxxa 21(13) (arcsin );1xx 21(15) (arctan );1xx 21(16) (arccot ).1xx 2(4).uu vuvvv( )( ).dydydy duf ug xdxdxdu dx或或,1111xxxxy.y21222xxy12xx1 y1212x)2( x112xx,1arctane2sin2xyx.y1arctan)(2xy ) (e2sin x2sine

6、x2cosxx221x1212xx2x21arctan2x2sinex2cosx2sinex112xx22sin2sin2earctan1earctan1xxyxx ,)(vuuvvuvu3sin (5 )1,.yxy求求11333221sin (5 )1 sin (5 )1 sin (5 )12yxxx解解2313sin (5 ) cos(5 ) (5 )2 sin (5 ) 1xxxx2315sin (5 ) cos(5 ).2 sin (5 ) 1xxx2313sin (5 ) sin(5 )2 sin (5 ) 1xxxxxxy)(21xxxxxxy)(211 (21xxxxxxx)2

7、11 (211 (21xxxxxx.812422xxxxxxxxxx)(sinnnnxfy)(sin)(sin1nnnnnxfxnfy)(sin)(sin1nnnxxn1cosnnnxx)(sin)(sin)(sin)(sincos1113nnnnnnnnnnxxfxxfxxn證證3）取得增量取得增量 u, v, 函函數(shù)數(shù) 也取得增量也取得增量 ( )( )u xyv x,()uuuv uu vyvvvv vv 00limlim()xxuvvuyxxyxv vv 故2( ) ( )( ) ( ).( )u x v xu x v xvx除法求導(dǎo)法則可簡(jiǎn)單地表示為除法求導(dǎo)法則可簡(jiǎn)單地表示為 2.u

8、u vuvvv當(dāng)當(dāng) x 取增量取增量 x 時(shí)時(shí), 函數(shù)函數(shù) u (x), v (x) 分別分別解：解：, )1,0(logaaxya那那么么),0(,yaxy)(logxa)(1ya 1aaylnxx1)ln(特別當(dāng)特別當(dāng)ea時(shí)時(shí),例例9. 求函數(shù)求函數(shù), )1,0(logaaxyaaxln1三、復(fù)合函數(shù)的求導(dǎo)法則三、復(fù)合函數(shù)的求導(dǎo)法則 定理定理3 設(shè)函數(shù)設(shè)函數(shù) u = g (x) 在點(diǎn)在點(diǎn) x 處可導(dǎo)處可導(dǎo), 函數(shù)函數(shù) y = f (u) 在點(diǎn)在點(diǎn) u = g (x) 處可導(dǎo)處可導(dǎo), 則復(fù)合函數(shù)則復(fù)合函數(shù) y = f (g(x)在點(diǎn)在點(diǎn) x 處可導(dǎo)處可導(dǎo), 且其導(dǎo)數(shù)為且其導(dǎo)數(shù)為 ( )( )

9、.dydydy duf ug xdxdxdu dx或或設(shè)設(shè) x 取增量取增量 x, 那么那么 u 取得相應(yīng)的增量取得相應(yīng)的增量 u, .yyuxux因?yàn)橐驗(yàn)?u = g (x) 可導(dǎo)可導(dǎo), 則必連續(xù)則必連續(xù), 所以所以 x 0 時(shí)時(shí), 000limlimlim,( )( ). 即即xuxyyudyf ug xxuxdx當(dāng)當(dāng) u = 0時(shí)時(shí), 可以證明上述公式仍然成立可以證明上述公式仍然成立. 從而從而 y 取得相應(yīng)的增量取得相應(yīng)的增量 y , 即即 u = g(x + x) g(x), y = f (u + u) f (u). u 0, 因而因而 當(dāng)當(dāng) u 0時(shí)時(shí), 有有證證中間變量的導(dǎo)數(shù)乘以

10、中間變量對(duì)自身變量的導(dǎo)數(shù)中間變量的導(dǎo)數(shù)乘以中間變量對(duì)自身變量的導(dǎo)數(shù). 設(shè)設(shè) y = f (u), u = g (v), v = h(x)都是可導(dǎo)函數(shù)都是可導(dǎo)函數(shù), 則復(fù)則復(fù)合函數(shù)合函數(shù) y = f (g(h(x) 對(duì)對(duì) x 的導(dǎo)數(shù)為的導(dǎo)數(shù)為 ( )( )( ).dydydy du dvf ug vh xdxdxdu dv dx或或公式表明公式表明, 復(fù)合函數(shù)的導(dǎo)數(shù)等于復(fù)合函數(shù)對(duì)復(fù)合函數(shù)的導(dǎo)數(shù)等于復(fù)合函數(shù)對(duì) 例例16 設(shè)設(shè) x 0, 證明冪函數(shù)的導(dǎo)數(shù)公式證明冪函數(shù)的導(dǎo)數(shù)公式 (x ) =x -1. 證證)()(lnxexxeln)ln(xxx1x22cossin( )cossincossinxxf xxxxx解解( )sincos ,fxxx sincos1.222f cos2( ),.cossin2xf xfxx求求2(sin ) cos lnsin (cos ) lnsin cos (ln ) yxxxxxxxxx解解2212coslnsinlnsin cosxxxxxxxsi