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习题231求函数的二阶导数(1)y2x2lnx(2)ye2x1(3)yxcosx(4)yetsint(5)22xay(6)yln(1x2)(7)ytanx(8)113xy(9)y(1x2)arctanx(10)xeyx(11)2xxey(12))1ln(2xxy解(1)xxy14214xy(2)ye2x122e2x1y2e2x124e2x1(3)yxcosxycosxxsinxysinxsinxxcosx2sinxxcosx(4)yetsintetcostet(costsint)yet(costsint)et(sintcost)2etcost(5)222222)(21xaxxaxay22222222222)(xaxaaxaxaxxxay(6)22212)1(11xxxxy222222)1()1(2)1()2(2)1(2xxxxxxy(7)ysec2xy2secx(secx)2secxsecxtanx2sec2xtanx(8)232233)1(3)1()1(xxxxy333433223)1()12(6)1(3)1(23)1(6xxxxxxxxxy(9)1arctan211)1(arctan222xxxxxxy212arctan2xxxy(10)22)1(1xxexexeyxxx3242)22(2)1(])1([xxxexxxexexeyxxxx(11))21()2(2222xexexeyxxx)23(24)21(222222xxexexxeyxxx(12)2222211)1221(11)1(11xxxxxxxxxyxxxxxxxxy1)1()12211)1(11222222设f(x)(x10)6f(2)?解f(x)6(x10)5f(x)30(x10)4f(x)120(x10)3f(2)120(210)32073603若f(x)存在求下列函数y的二阶导数22dxyd(1)yf(x2)(2)yln[f(x)]解(1)yf(x2)(x2)2xf(x2)y2f(x2)2x2xf(x2)2f(x2)4x2f(x2)(2))()(1xfxfy2)]([)()()()(xfxfxfxfxfy22)]([)]([)()(xfxfxfxf4试从ydydx1导出(1)322)(yydyxd(2)5233)()(3yyyydyxd解(1)3222)(1)(11yyyyydydxydxdydyddydxdyddyxd(2)dydxyydxdyydyddyxd333352623)()(31)()(3)(yyyyyyyyyyy5已知物体的运动规律为sAsint(A、是常数)求物体运动的加速度并验证0222sdtsd解tAdtdscostAdtsdsin22222dtsd就是物体运动的加速度0sinsin22222tAtAsdtsd6验证函数yC1exC2ex(C1C2是常数)满足关系式y2y0解yC1exC2exyC12exC22exy2y(C12exC22ex)2(C1exC2ex)(C12exC22ex)(C12exC22ex)07验证函数yexsinx满足关系式y2y2y0解yexsinxexcosxex(sinxcosx)yex(sinxcosx)ex(cosxsinx)2excosxy2y2y2excosx2ex(sinxcosx)2exsinx2excosx2exsinx2excosx2exsinx08求下列函数的n阶导数的一般表达式(1)yxna1xn1a2xn2an1xan(a1a2an都是常数)(2)ysin2x(3)yxlnx(4)yxex解(1)ynxn1(n1)a1xn2(n2)a2xn3an1yn(n1)xn2(n1)(n2)a1xn3(n2)(n3)a2xn4an2y(n)n(n1)(n2)21x0n!(2)y2sinxcosxsin2x)22sin(22cos2xxy)222sin(2)22cos(222xxy)232sin(2)222cos(233)4(xxy]2)1(2sin[21)(nxynn(3)1lnxy11xxyy(1)x2y(4)(1)(2)x3y(n)(1)(2)(3)(n2)xn1112)!2()1()!2()1(nnnnxnxn(4)yexxexyexexxex2exxexy2exexxex3exxexy(n)nexxexex(nx)9求下列函数所指定的阶的导数(1)yexcosx求y(4)(2)yxshx求y(100)(3)yx2sin2x求y(50).解(1)令uexvcosx有uuuu(4)exvsinxvcosxvsinxv(4)cosx所以y(4)u(4)v4uv6uv4uvuv(4)ex[cosx4(sinx)6(cosx)4sinxcosx]4excosx(2)令uxvshx则有u1u0vchxvshxv(99)chxv(100)shx所以)100()99(99100)98(98100)98(2100)99(1100)100()100(vuvuCvuCvuCvuCvuy100chxxshx(3)令ux2vsin2x则有u2xu2u0xxv2sin2)2482sin(24848)48(v(49)249cos2xv(50)250sin2x所以)50()49(4950)48(4850)48(250)49(1150)50()50(vuvuCvuCvuCvuCvuy)50()49(4950)48(4850vuvuCvuC)2sin2(2cos22502sin22249505024928xxxxx)2sin212252cos502sin(2250xxxxx
本文标题:2-3高等数学同济大学第六版本
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