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时间:2019-11-22
《经济数学基础平时作业》由会员上传分享,免费在线阅读,更多相关内容在工程资料-天天文库。
1、经济数学基础平时作业作业(一)(一)填空题.答案:0(・x-sinxl.limxtO2•设/(%)=V+1,2、)U(—2,+8)C.(-00^-2)U(-2J)U(l,+oo)D.(—8厂2)kJ(―2,+°°)或(—8,1)LJ(1,+°°)1.当XT+oo时,下列变量为无穷小量的是(D)7X"B.x+12.下列极限说傅正确的是(A.ln(l+x)C.exsinxD.xB.lim—=1XTO+XC.limxsin—=1«vtOxDlnn沁=1"TooX3.设y=lg2x,贝ijdy=)•A.—dxB.—-—2xxlnlOdxD.-dxXC.函数/(兀)在点xo处连续5•当兀T0111卜•列变量是无穷小量的是(C4.3、若函数.f⑴在点皿处可导,贝ll(A.函数金)在点勺处有定义)是鹏的.B.lim/(x)=A,但AH/(x0)XT%D.函数/(兀)在点xo处可微)・寸+xe+HE叱jUH一nUH一xxQ+XF—x+—II“C“8TH(寸)(I+x—m)。!(I+x—『)x。!u+n。1X。!^^—ImT,un〒、丿unTui二7HI二(I+x丄、)(14、『)IIY—『e寸—e寸—YFY(寸Ix)(zIdZTX8+H9IH2TX、J217(E—X)(z—X)9+y」—h・・OTKInHLU二I+II+VTH一一一二H、IT75、—xHsoo・a(E)(z)(I+x)(-—x)II13—I—HIE二e+XE—H二卫i±・I®S0I.<(I))H(X)、J亘"XH(T)J^sh+duiu寸II0ah-OHHOVKHlns6X,g+—UISKI(zIx)lnsz!I(zIx)lns(zIY)lnsz!Jdllx^isNng-H~P7~宜ZTKz!、(9)x「ozSIS忘ins•「◎V』UISOTXYulinsOTX—HIX—H■XIH—X・OU一nElmHECxm二xmws(s)E—O+O+E7—o+o—eHH寸+Z+E82寸+XZ+LE86、T”vYUn-n0uniezS+VEJ遵二(寸)答案:(1)因为/(兀)在兀=0处有极限存在,贝ij有乂lim/(x)=lim(xsin—+b)=bxt(fxt(fxlimf(x)=lim也匕=1XT()・XT()・X即b=l所以当a为实数、b=l吋,于⑴在x=0处极限存在.(2)因为/(对在兀=0处连续,贝怖lim/(x)=lim/(x)=/(O)xt(eA->0+又f(O)=af结合(1)可知g二b=l所以当a=b=l时,f(x)在x=0处连续.3.计算卜•列函数的导数或微分:(1)y=x丄7、3--答案:/8、=[(3x-5)2y=_l(3x-5)2(3x-5)z=--(3x-5)2(4)y=4x-xex,求;/丄1-丄答案:y/=(x2y-(xexy=-x2-ex-xex(5)y=eaxsinbx,求dy答案:)/=(eaxYsinbx-eax(sinbx)'=eax(axysinbx-eaxcosbx(bxy=aeaxsinbx-beaxcosbxdy=ydx-(aeaxsinbx-beaxcosbx)dx+2V+log2x-22,求答案:yz=2x+2'In2xln2⑵y=^f求讨cx+d答案:/=(ax+b9、(cx+d)-(ax+b)(cx+d_a(cx+d)-(ax+b)c(cx+d)?(cx+d)2ad-he(cx+d)2(3)y二/1:,求)/y/3x—5丄310、~r+—xx22(6)y=ex+xy[x,求dy答案:=(/Y+(丿y=*(丄y+3兀》x2px31dy=ydx-(——r+—兀?)dxx2(7)y=cosVx-e"r,求dy-x答案:)/=(cosVx)/-(e~v=-sinVx(Vx)z-e~r(-x2)'=-血二+2xe~r2』xg*—吓)dx2』x(8)y=sinHx+sinnx,求)/答案11、:)/=[(sin%)"]"+(sinmx),=7?(sinx),,_l(sinx)z+cos/ix(nx)z=n(sinx)0"1cosx+ncos/ix(9)y=ln(x+Jl+〒),求;/11ii-1答案:/=^7^(x+vr^)/=7^?(1+i(1+%2)22x)_](]+乂)_]X+Jl+兀2Jl+Jl+兀~/“、nc°l;.l+VP~-V2x+/(10)y=2*,求y・1
2、)U(—2,+8)C.(-00^-2)U(-2J)U(l,+oo)D.(—8厂2)kJ(―2,+°°)或(—8,1)LJ(1,+°°)1.当XT+oo时,下列变量为无穷小量的是(D)7X"B.x+12.下列极限说傅正确的是(A.ln(l+x)C.exsinxD.xB.lim—=1XTO+XC.limxsin—=1«vtOxDlnn沁=1"TooX3.设y=lg2x,贝ijdy=)•A.—dxB.—-—2xxlnlOdxD.-dxXC.函数/(兀)在点xo处连续5•当兀T0111卜•列变量是无穷小量的是(C4.
3、若函数.f⑴在点皿处可导,贝ll(A.函数金)在点勺处有定义)是鹏的.B.lim/(x)=A,但AH/(x0)XT%D.函数/(兀)在点xo处可微)・寸+xe+HE叱jUH一nUH一xxQ+XF—x+—II“C“8TH(寸)(I+x—m)。!(I+x—『)x。!u+n。1X。!^^—ImT,un〒、丿unTui二7HI二(I+x丄、)(1
4、『)IIY—『e寸—e寸—YFY(寸Ix)(zIdZTX8+H9IH2TX、J217(E—X)(z—X)9+y」—h・・OTKInHLU二I+II+VTH一一一二H、IT7
5、—xHsoo・a(E)(z)(I+x)(-—x)II13—I—HIE二e+XE—H二卫i±・I®S0I.<(I))H(X)、J亘"XH(T)J^sh+duiu寸II0ah-OHHOVKHlns6X,g+—UISKI(zIx)lnsz!I(zIx)lns(zIY)lnsz!Jdllx^isNng-H~P7~宜ZTKz!、(9)x「ozSIS忘ins•「◎V』UISOTXYulinsOTX—HIX—H■XIH—X・OU一nElmHECxm二xmws(s)E—O+O+E7—o+o—eHH寸+Z+E82寸+XZ+LE8
6、T”vYUn-n0uniezS+VEJ遵二(寸)答案:(1)因为/(兀)在兀=0处有极限存在,贝ij有乂lim/(x)=lim(xsin—+b)=bxt(fxt(fxlimf(x)=lim也匕=1XT()・XT()・X即b=l所以当a为实数、b=l吋,于⑴在x=0处极限存在.(2)因为/(对在兀=0处连续,贝怖lim/(x)=lim/(x)=/(O)xt(eA->0+又f(O)=af结合(1)可知g二b=l所以当a=b=l时,f(x)在x=0处连续.3.计算卜•列函数的导数或微分:(1)y=x丄
7、3--答案:/
8、=[(3x-5)2y=_l(3x-5)2(3x-5)z=--(3x-5)2(4)y=4x-xex,求;/丄1-丄答案:y/=(x2y-(xexy=-x2-ex-xex(5)y=eaxsinbx,求dy答案:)/=(eaxYsinbx-eax(sinbx)'=eax(axysinbx-eaxcosbx(bxy=aeaxsinbx-beaxcosbxdy=ydx-(aeaxsinbx-beaxcosbx)dx+2V+log2x-22,求答案:yz=2x+2'In2xln2⑵y=^f求讨cx+d答案:/=(ax+b
9、(cx+d)-(ax+b)(cx+d_a(cx+d)-(ax+b)c(cx+d)?(cx+d)2ad-he(cx+d)2(3)y二/1:,求)/y/3x—5丄3
10、~r+—xx22(6)y=ex+xy[x,求dy答案:=(/Y+(丿y=*(丄y+3兀》x2px31dy=ydx-(——r+—兀?)dxx2(7)y=cosVx-e"r,求dy-x答案:)/=(cosVx)/-(e~v=-sinVx(Vx)z-e~r(-x2)'=-血二+2xe~r2』xg*—吓)dx2』x(8)y=sinHx+sinnx,求)/答案
11、:)/=[(sin%)"]"+(sinmx),=7?(sinx),,_l(sinx)z+cos/ix(nx)z=n(sinx)0"1cosx+ncos/ix(9)y=ln(x+Jl+〒),求;/11ii-1答案:/=^7^(x+vr^)/=7^?(1+i(1+%2)22x)_](]+乂)_]X+Jl+兀2Jl+Jl+兀~/“、nc°l;.l+VP~-V2x+/(10)y=2*,求y・1
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