同济高数课后习题答案解析.pdf

同济大学高等数学同济大学高等数学 一、求下列极限一、求下列极限1 1 1 1、、sin ()sin ()sin ()sin ()limlimlimlim x x x xx x x x x x x x→→→→− − − − − − − −2 22 21 11 1 1 1；；解一：解一：()()12sin1 cos1lim02xxx x→−−==原式解二解二： ：()()11sin1sin1limlim011xxxxxx→→−−==−+原式2 2 2 2、、limlimlimlimsinsinsinsinx x x xx x x x x x x x→→→→2 22 20 03 3 解一：解一：00021311limlimlim6sin3 cos39sin3cos39xxxxx xxxx→→→==⋅=原式解二：解二：sin 3 ~30021limlim6sin 3 cos39 cos39xxxxxx xxxx→→===原式3 3 3 3、、2 2 2 20 0 0 0tan2tan2tan2tan2limlimlimlimsin 3sin 3sin 3sin 3x x x xxxxxxxxx x x x x→→→→解：解：()2tan2 ~2 ,sin3 ~32022lim93xxxxxxx→=原式=4 4 4 4、、0 0 0 0limlimlimlimln(1)ln(1)ln(1)ln(1)x x x xx x x x x x x x→→→→+ + + +解一：解一：()001limlim 111 1xxxx→→==+=+原式解二：解二：()1011lim1lnln 1x xex→=== +原式5 5、、2 2 2 2limlimlimlimx x x xx x x xx x x x x x x x→∞→∞→∞→∞− − − −⎛⎞⎛⎞⎛⎞⎛⎞ ⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎝⎠解一：解一：()2222lim 1xxex− ⋅ − −→∞⎛⎞=−=⎜⎟⎝⎠原式解二：解二：() 1211 ln2ln22limlimln2lim22limxxxxxxxxxxxxxx xeeeee−−→∞→∞→∞−−−−− −→∞− −−=====原式6 6 6 6、、( ( ( () ) ) )1 1 1 1 1 1 1 1 1 1 1 1lim 32lim 32lim 32lim 32x x x x x x x xx x x x− − − − →→→→− − − −解一：解一：()()11222 0lim 1 2t x t tte= −−⋅ −−→=−=令 原式解二：解二：1( 2)2 211 222 21lim[1 (22 )]{lim[1 (22 )]}xxxxxxe−−→−−−→=+−=+−=i原式7 7 7 7、、3 3 3 30 0 0 0sinsinsinsinlimlimlimlim x x x xxxxxxxxx x x x x→→→→− − − −解：解：2001 cossin1limlim366xxxx xx→→−===原式8 8 8 8、、1 1 1 111111111limlimlimlim1ln1ln1ln1lnx x x xxxxxxxxx→→→→⎛⎞⎛⎞⎛⎞⎛⎞− − − −⎜⎟⎜⎟⎜⎟⎜⎟− − − −⎝⎠⎝⎠⎝⎠⎝⎠解：解：111111ln11limlimlim1(1)lnln1ln11limln1 12xxxxxxxx xxxxxxxxx→→→→−−+−===−−+−+−== −+ +原式9 9 9 9、、12121212limlimlimlim22222222n n n nnnnnnnnn n n n n→∞→∞→∞→∞++++++++++++⎛⎞⎛⎞⎛⎞⎛⎞− − − −⎜⎟⎜⎟⎜⎟⎜⎟+ + + +⎝⎠⎝⎠⎝⎠⎝⎠⋯⋯⋯⋯解：解：()()221122limlim222 21lim422nnnn nnnnnn nnn n→∞→∞→∞⎛⎞+⎜⎟+−−=−=⎜⎟++⎜⎟⎝⎠ −== −+原式10101010、、3 3 3 3 2 2 2 20 0 0 0 9 9 9 90 0 0 0sinsinsinsin limlimlimlimx x x xx x x xt dtt dtt dtt dtx x x x→→→→∫ ∫ ∫ ∫解：解：26686003sin1sin1limlim933xxxxx xx→→===原式11111111、、0 0 0 02 2 2 2arctanarctanarctanarctan limlimlimlim 1 1 1 1x x x xx x x xtdttdttdttdtx x x x→+∞→+∞→+∞→+∞+ + + +∫ ∫ ∫ ∫ 。

)12 222arctan1limlim arctan11122 1lim arctanlim1122xxxxxxxxxxxππ→+∞→+∞−→+∞→+∞==⋅+ +⋅=⋅+ =⋅ =解：原式二、求下列导数或微分二、求下列导数或微分1 1 1 1、、设设4 4 4 4tan1tan1tan1tan1yxxxyxxxyxxxyxxx=−+=−+=−+=−+，求，求dydydydy解一：解一：dyydx′=()324tansecxxxx dx=−−解二解二： ：()34tantandyx dxxdxxdx=−+()32324tansec4tansecx dxxdxxxdxxxxx dx=−− ⋅=−−2 2 2 2、、设设2 2 2 2 1cos1cos1cos1cosx x x x y y y yx x x x= = = =+ + + +，求，求y y y y′ ′ ′ ′()() ()() ()222 ln2 1 cos2sin1 cos2sincosln2ln21 cosxxxxxy xxxx⋅+−−′ =++⋅+= +解:3 3 3 3、、设设5 5 5 51 1 1 1log sinlog sinlog sinlog siny y y yx x x x= = = =，求，求y y y y′ ′ ′ ′()221cos111cos11ln5sinln5sinxyxxxxx−′ =⋅⋅ −= −⋅ ⋅解 4 4 4 4、、设设2 2 2 2ln(1)ln(1)ln(1)ln(1)yxxyxxyxxyxx=++=++=++=++，求，求y y y y′′ ′′ ′′ ′′()()()()1 22 22221 2233 222211112211111111212yxx xxxxxxxxyxxxx−−−−⎛⎞′/=⋅++⋅⎜⎟/++⎝⎠++=⋅ +++=+′′= −+⋅= −+解：5 5 5 5、、设设lnlnlnlnxyyxyyxyyxyy=+=+=+=+，求，求dydydydy dxdxdxdx 111yyyyyy′′′=+⋅∴=+解：6 6 6 6、、设设sin()0sin()0sin()0sin()0x x x xyexyyexyyexyyexy+=+=+=+=，求，求dydydydy dxdxdxdx () ()()()()cos0coscosxxxxyeyexyy xyy exyyexxy′′++⋅+=+′∴=−+解7 7 7 7、、设设2 2 2 21 1 1 1 arctanarctanarctanarctanxtxtxtxt ytytytyt⎧ ⎧ ⎧ ⎧⎪ ⎪ ⎪ ⎪=+=+=+=+⎨ ⎨ ⎨ ⎨= = = =⎪ ⎪ ⎪ ⎪⎩ ⎩ ⎩ ⎩，求，求1 1 1 1t t t tdydydydy dxdxdxdx= = = =()()1 2221 2211 11 1122 22ttdyt dxtttdy dx−−=++== +⋅∴=解8 8 8 8、、设设2 2 2 2ln(1)ln(1)ln(1)ln(1)arctanarctanarctanarctanxtxtxtxtyttyttyttytt⎧ ⎧ ⎧ ⎧=+=+=+=+⎨ ⎨ ⎨ ⎨= −= −= −= −⎩ ⎩ ⎩ ⎩，求，求2 2 2 22 2 2 2d yd yd yd y dxdxdxdx222222111 12211212 1421dytt dxtt td d ytdt dxdxttdtt−+== ⋅+ ⎛⎞⎜⎟⎝⎠+∴=== ⋅+解9 9 9 9、、设设( ( ( ( ) ) ) )tantantantan(0)(0)(0)(0)x x x xyxxyxxyxxyxx=>=>=>=>，求，求y y y y′ ′ ′ ′()22tan ln2lntan ln 11seclntan1seclntanseclntanxxyxxyxxxyxyyxxxxexxxxx=′ =⋅+⋅⎛⎞′ =⋅+⋅⎜⎟⎝⎠+=解：10101010、、设设4 4 4 45 5 5 52(3)2(3)2(3)2(3) (1)(1)(1)(1)xxxxxxxxy y y yx x x x+−+−+−+−= = = =+ + + +，求，求y y y y′ ′ ′ ′()()()()()()()() ()451lnln24ln 35ln 12 1145 2231145 22312 314524311yxxxyyxxxyyxxxxxxxxx=++−−+′ =−−+−+⎛⎞′ =−−=⎜⎟⎜⎟+−+⎝⎠+−⎛⎞−−⎜⎟+−+⎝⎠+解11111111、、设设2 2 2 23 3 3 3sinsinsinsinx x x xt t t tx x x xye dtye dtye dtye dt= = = =∫ ∫ ∫ ∫，求，求y y y y′ ′ ′ ′32226sin00sin2cos3xxttxxye dte dtyexx e=−′∴=−∫∫解12121212、、设设3 3 3 31 1 1 1sin0sin0sin0sin0( )( )( )( ) 00000000xxxxxxxxf xf xf xf xx x x x x x x x⎧ ⎧ ⎧ ⎧≠ ≠ ≠ ≠⎪ ⎪ ⎪ ⎪= = = =⎨ ⎨ ⎨ ⎨ ⎪ ⎪ ⎪ ⎪= = = =⎩ ⎩ ⎩ ⎩，求，求( )( )( )( )fxfxfxfx′ ′ ′ ′()( )()( )323223002020 111(sin)3sincos113sincos0 1sin000limlim1limsin0113sincos,00,0xxxxyxxxxxxxxxxx xxfxfxyxxxxxxxyxx x−∆ →∆ →∆ →≠′′==+⋅ −=−=∆+∆−∆′==∆∆ ⎛⎞=∆=⎜⎟∆⎝⎠ ⎧−≠⎪′∴=⎨ ⎪=⎩解，，三、求下列积分三、求下列积分1 1 1 1、、x x x xedxedxedxedx− − − −∫ ∫ ∫ ∫()xxe dxec−−= −−= −+∫解：原式2 2 2 2、、2 2 2 2x x x xx x x xe e e edxdxdxdx∫ ∫ ∫ ∫()2 2ln221ln 2xxxxe edxceec⎛⎞⎜⎟⎛⎞⎝⎠==+⎜⎟⎛⎞⎝⎠⎜⎟⎝⎠=+−∫解 ： 原 式3 3 3 3、、csccsccsccscdxdxdxdxx x x x∫ ∫ ∫ ∫1 1sincosxdxxc== −+∫解：原式4 4 4 4、、2 2 2 2x x x xxedxxedxxedxxedx∫ ∫ ∫ ∫22211 22xxe dxec==+∫解：原式5 5 5 5、、3 3 3 3(ln )(ln )(ln )(ln )x x x xdxdxdxdxx x x x∫ ∫ ∫ ∫()()341lnlnln4x dxxc==+∫解:原式6 6 6 6、、xxxxxxxxdxdxdxdx eeeeeeee− − − −+ + + +∫ ∫ ∫ ∫()22arctan11xx x xxededxecee===+++∫∫解：原式7 7 7 7、、4 4 4 42 2 2 21 1 1 1x x x xdxdxdxdxx x x x+ + + +∫ 。