大學微積分導數(shù)的計算ppt課件

1、導數(shù)定義式導數(shù)定義式xxfxxfx)()(lim000)( 0 xf。00)()(lim0 xxxfxfxx).1 ( , 21)(lim1)(11fxxfxxfx求處連續(xù)，且在設).( , 1)0( ,2)()()( 2xffhxhfxfhxfx,h求且，有已知對任意的實數(shù)練習練習, 10 ,11, 01),1ln()(11. 3 . 2xxxxxxf設函數(shù)例處的連續(xù)性與可導性。在討論0)(xxf解：連續(xù)性)0() 1 (f)01ln( 0 )00()2(f0limx)1ln(x0 )00(f0limx)11(xx00)00()00(ff0)(lim0 xfx),0(0)(lim)3(0fx

2、fx.0)(處連續(xù)在 xxf, 10 ,11, 01),1ln()(11. 3 . 2xxxxxxf設函數(shù)例處的連續(xù)性與可導性。在討論0)(xxf解：可導性)0() 1 (f00)1ln(xx)0()2(f0limx),0()0(ff.0)(處不可導在 xxf0limxxx)1ln( 10limx00)11(xxx0limxxx111211 , 21, 10, 1)(13. 3 . 22xxbaxxxxf在設例的值。連續(xù)且可導，求ba,解：點連續(xù)函數(shù)在1x)01 ()01 (ff1limx) 1(2x1limx)(bax0ba ab1 , 21, 10, 1)(13. 3 . 22xxbaxx

3、xxf在設例的值。連續(xù)且可導，求ba,1點可導函數(shù)在 x),1 () 1 (ff1limx1limx10) 1(2xx10)(xbax21limx1xaaxa2b導導( (函函) )數(shù)定義式數(shù)定義式)( xfxxfxxfx)()(lim02.2 導數(shù)的基本公式與求導法則導數(shù)的基本公式與求導法則 2.2.1 基本初等函數(shù)的導數(shù)的導數(shù)。求xxfln)(. 1解 )(ln xxxxxxln)ln(lim0)1ln(1lim0 xxxx)1ln(1lim0 xxxxxxxxxxxx)1ln(1lim0exln1x1xx1 )(ln公式：的導數(shù)。求xxfsin)(. 2解 )(sin xxxxxxsin

4、)sin(lim0 xxxxx2sin)2cos(2lim022sinlim)2cos(lim00 xxxxxxxcos。）（公式xxcossin:cos ）（xxsin2.2.2 函數(shù)的和、差、積、商的求導法則處可導，則都在與設函數(shù)定理xxvxu)()(1 . 2 . 2；)( )( )()() 1 (xvxuxvxu；)( )()()( )()()2(xvxuxvxuxvxu；)( )()3(xcuxcu。，0)()()( )()()( )()()4(2xvxvxvxuxvxuxvxu。，特別，0)()()( )(12xvxvxvxv；)( )()()( )()()2(xvxuxvxuxv

5、xu證明：)()(xvxuxxvxuxxvxxux)()()()(lim0 xxvxuxxvxuxxvxuxxvxxux)()()()()()()()(lim0 xxvxxvxuxxvxuxxux)()()()()()(lim0。，0)()()( )()()( )()()4(2xvxvxvxuxvxuxvxu )(1xv分析：)(1)()()(xvxuxvxu)(1)()(1)( xvxuxvxuxxvxxvx)(1)(1lim0注：（1加、減、乘法則可推廣到有限個函數(shù)的情況。；如：)( )( )( )()()(2121xuxuxuxuxuxunn；)( )()()()( )()()()( )

6、()()(xwxvxuxwxvxuxwxvxuxwxvxu；)( )( )( )()()()2(22112211xukxukxukxukxukxuknnnn處可導。均在為常數(shù)，其中xxuxukkknn)(,),(,121的導數(shù)。，求例) 10(logf(x)2.2.1xaa )log(xa解。axln1 )lnln(ax )(lnln1xaaxln1 )log(xa公式：的導數(shù)。求例anxtf(x)2.2.2解 )tanx(cossinxxxxxxx2cos )(cossincos )(sinxxx222cossincosx2secx2sec )tanx(公式：x2csc )cotx(x2co

7、s1的導數(shù)。求例xsecf(x)2.2.3解cos1x )secx(xx2cos )(cosxx2cossinsecxtanxxxtansec )secx(xxccotcsc )scx(練習：求下列函數(shù)的導數(shù)x2siny (1)xx2)1(y (2)baxy (3)babaxy (4)xalogy (5)2.2.3 反函數(shù)的導數(shù)定理定理2.2.2 )(),()(xfyfxxfy若的反函數(shù)為設10)( )( 1)(1xfxfyf，0)()(1)( 11yfyfxf，或即互為反函數(shù)的兩個函數(shù)，它們的導數(shù)互為倒數(shù)。即互為反函數(shù)的兩個函數(shù)，它們的導數(shù)互為倒數(shù)。處可導，且在對應點則處可導，且在yyfxx

8、fx)(,)(10由導數(shù)定義得)(1yfyyfyyfy)()(lim110yxy0lim. 00 xy時，故當)( 1xfxyy1lim0 xyx1lim0處連續(xù)，在對應點連續(xù)性知由反函數(shù)的連續(xù)可導，故證：因為yyfxxfxf)(,)()(1的導數(shù)。求例arcsinxy2.2.4，的反函數(shù)為解)22(sinarcsinxy yyxarcsinx)( )(sin1yycos1y2sin11211x，211 )(arcsinxx211 )(arccosxx的導數(shù)。求例arctanxy2.2.5，的反函數(shù)為解yxtanarctanxy arctanx)( )(tan1yy2sec1y2tan1121

9、1x，211 )(arctanxx。211 )cot(xxarc的導數(shù)。，求例) 10(ay2.2.6xa，的反函數(shù)為解)0(logay yxyax )a (x )(log1yaayln11aylnaaxlnaaxxln )a ( )e (xxe例2.2.7 求下列函數(shù)的導數(shù)：xxxxln12siny (1)3解3ln12siny (1)xxxx67)(ln)12(sin)(xxxx1067167xx16761xx arcsiny (2)3解3)arcsin(y xx33)(arcsinarcsin )(xxxx2321arcsin3xxxx21arctany (3)xx解2)1arctan(

10、yxx2222)1()(arctan )1()1( )(arctanxxxxx2222)1()(arctan2()1(11xxxxxxxxxarctan)1(211224作業(yè)作業(yè)P63 6(1-10)P63 6(1-10)P64 10P64 10P65 15(1,3),16P65 15(1,3),16；)( )()()( )()()2(xvxuxvxuxvxu證明：)()(xvxuxxvxuxxvxxux)()()()(lim0 xxvxuxxvxuxxvxuxxvxxux)()()()()()()()(lim0 xxvxxvxuxxvxuxxux)()()()()()(lim0)()()(lim)()()(lim00 xuxxvxxvxxvxxuxxuxx；)( )()()( xvxuxvxu。，0)()()( )()()( )()()4(2xvxvxvxuxvx