2021年國開電大《高等數(shù)學(xué)基礎(chǔ)》第二次作業(yè)答案

2021年國開電大《高等數(shù)學(xué)基礎(chǔ)》第二次作業(yè)答案高等數(shù)學(xué)基礎(chǔ)第二次作業(yè)（一）單項選擇題

第3章導(dǎo)數(shù)與微分f(00且極限

f(x存在，則

f(x)．x0 x

x0 xA.f(0)A.f(0)B.fC.f(x)D.0f(xx0

可導(dǎo)，則h0

f(x0

2h)f(x02h

（．A. 2f(x) B. f(x)0 0C. 2f(x) D. f(x)0 0fx)ff(x)exlimx0 x

（．A. e B. 2eC. 1e D. 1e2 4設(shè)f(x)x(x)(x2) (x99)，則f()（．A.99B.99C.99!D.99!⒌下列結(jié)論中正確的是（C．f(xxx可導(dǎo)．0 0f(xxx可導(dǎo)．0 0f(xxx有極限．0 0f(xxx連續(xù)．0 0（二）填空題 1f(xx2sinx0,

, x0，則f(0) 0 ．x0f(exe2xx

df(lnx)2lnx5dx xf(x

1在2)處的切線斜率是 1．x2xπ⒋曲線f(x)sinx在( ,處的切線方程是y1．2⒌設(shè)yx2x，則y x2xlnx⒍設(shè)yxlnx，則y 1．x（三）計算題⒈求下列函數(shù)的導(dǎo)數(shù)y：x⑴y(x xx3 3 1 3y(x2ex3ex) x2ex2

x2ex3ex3 3 1=ex(x

x22⑵ycotxx2lnxy

cosx sinxsinxcosxcosx x2x2lnx)( 2xlnx )sinx= 1 2xlnx

sin2x x⑶y

sin2xx2lnxy2xlnxxx(2lnxln2x ln2x⑷y

cosx2xx3y

(sinx2xln2)x3(cosx2x)3x2x6xsinxln22xx3cosx32x=x4⑸y

lnxxsinx1( 2x)sinxcosx(lnxx2)1解：yx sin2x2x2)sinxxcos(lnxx2)=xsin2x⑹yx4sinxlnxsinxy4x3

(cosxlnx )x=4x3

cosxlnx

sinxx⑺y

sinxx23xy

(cosx2x)3x3xlnxx2)32x=⑻yex

cosx2xlnxx23xtanxlnx解：y(ex

tanx

ex )1cos2x xex(sinxcosx1= cos2x x⒉求下列函數(shù)的導(dǎo)數(shù)y：x⑴yex2 xxx解：yex 1 e2 xxx2x⑵ylncosx解：ysinxtanxcosxx x x117⑶yx x x1171yx1

x

x

x8 7 1所以yx88⑷ysin2xy2sinxcosxsin2x12 x1 1 212 x1所以y

(xx2)3 )3⑸ysinx2ycosx22x2xcosx2⑹ycosexysinexex=exsinex⑺ysin

xcosnxy(sin

x)cosnxsin

x(cosnx)=nsinn1xcosxcosnxsinnx(sinnx)n=nsinn1x(cosxcosnxsinxsinnx)⑻y5sinx解：設(shè)y5u usinxyyuu x

=

ln5cosxln55sin

cosx⑼yecosxyeu

ucosxyyuu x

=eu

(sinx)ecosxsinx⒊在下列方程中， 是由方程確定的函數(shù)，求 ：解：將方程兩邊對x求導(dǎo)：ycosxysinx=2e2yy移項 y(cosx2e2y)ysinx所以：y ysinxcosx2e2y⑵ycosylnx解：將方程兩邊對x求導(dǎo)：y(cosy)lnxcosy(lnx)ysinyylnx

cosyx移項sinylnx所以：y cosy

cosyxlnxsiny)⑶2xsinyx2y

2xyx2y' 2x x2解：2simy2xcosyy'2x2simy

y'y2 y y2y' y 2xcosyx2y2

2xy2y2simy2xy2cosyx2⑷yxlnyy1yy解得y 1y1⑸lnxeyy2解：將方程兩邊對x求導(dǎo)：1 eyy2yyx1整理得：y 1x(2yey)⑹y21exsiny解：將方程兩邊對x求導(dǎo)：2yyexsinyexcosyyy

exsiny2yexcosy⑺eyexy3解：將方程兩邊對x求導(dǎo)：eyyex3y2y整理得：yey

ex3y2⑻y5x2y解：將方程兩邊對x求導(dǎo)：y5xln52yln2yy

5xln512yln⒋求下列函數(shù)的微分dy：⑴ycotxcscx解：因為y

1 (

) 1

cosxsin2x sinx sin2x sin2x=1cosxsin2x所以 dy

1cosxdx⑵y

lnxsinx

sin2x1sinxcosxlnx解：因為yxsin2xsinxxcosxlnx= xsin2xsinxxcosxlnx所以dy= dxxsin2x⑶ysin2x解：設(shè)yu2, usinx則yyuu x=2ucosx2sinxcosx=sin2x所以dy=sin2xdx⑷ytanex解：: ytanu, uex則yyuu x1=cos2u

ex=所以dy=

excos2exexdxcos2ex⒌求下列函數(shù)的二階導(dǎo)數(shù)：x⑴yx12 x解：y12 x12 x( )1 12 xy x24⑵y3xy3xln3y(3xln3)3xln3ln3=⑶ylnxyy

1x(1)1x x2⑷yxsinx解：ysinxxcosxy(sinxxcosx)cosxc