一、导数的运算1, 已知2211xxy,则 y =()。A, )1(22xx B, 2)1(4xx C, 22)1(2xx D, 22 )1(4xx解22 )1(4xx。2 xxycos22,则 y =()。A, xxxxxcossin2cos42 B, xxxxx2cossin2cos4 C, xxxxx22cossin2cos4 D, xxxx22cossin2cos4解)cos/2(2xxyxxxxx22cossin2cos4。3 2sin xy,则 y =()。A, 2cosx B, 2cos2xx C, 2cos2x D, xxcos2令2xu,则uysin,uyucos,xu x2,所以xuxuyy2cos2xx。4 )1ln(2xxy,则 y =()。A, 211x B, 211xx C, 212xx D, 212xx令 y=lnu,21vxu,v=1+x2 则uyu1,121211vuv,xvx2所以xvuxvuyy211x。今后可约定yy x,省略下x标。5 3)sin(ln xy,则 y =()。A, 23)(ln)cos(lnxx B, 3)cos(ln xC, )(ln)cos(ln33xxx D, 23)(ln)cos(ln3xxx令vysin, 3uv, xuln,,则xuvuvyy23)(ln)cos(ln3xxx。6:xy2sin3, 则 y =()。A, xx2sin32cos2 B, 3ln2cos2xC, 3ln32cos22sinxx D, 3ln32cos2sinxx解3ln22cos32sinxyx3ln32cos22sinxx。7, 设函数)2arccos( xey,则dxdy等于()A.2)2arccos(41xex B.2)2arccos(412xex C .2)2arccos(212xex D .2)2arccos(12xex解答:)()2arccos( xey=])2[arccos()2arccos(xex=)2(412)2arccos(xxex=2)2arccos(412xex8,导数是31x的函数是 ( ) A,3212x B ,2414x C ,1414x D ,4212x解答 :)321(2x=)(212x=x-3 9,函数31x的导数是 ( ) A,23x B ,43x C ,23x D ,43x解答 :)1(3x=)(3x=-3x-410, 设xy2sin,则 y =()。A, 2x2sin B, 2xdx2sin C, x2sin D,xdx2sin11, 设xyln1,则y = ()A, xxln11 B, xxln121 C, xln11 D, xln1212, bxeyax sin,则 y = ()A, )sincos(bxabxbeax B, )sincos(bxabxbeax C, )sincos(bxabxbeax D, )sincos(bxabxbeaxbxbxdeaxsin)(sin·axdeaxe·bxbxbxdsin)(cos·)( axdeaxdxbxabxbeax)sincos(。13,)(21x=(), A, 2121 x B, 2121 x C, 2321 x D,2321 x14,)(2xe = ()A, xe2 B, xe2 C, xe22 D,dxex2215,)(log 2 x = ()A, ex2log2 B, ex2log1 C, dxx1 D,x116,)5(x = ()A, 5ln5x B, x5 C, dxx5 D,5ln517,)2(lnx = ()A, Lnx B, x2 C, x21 D,x118, 设 y=sin7x , 则 y = ()A,-7cos7x B, 7cosx C, 7cos7x D, cos7x 19, 设 y = xcos(-x) ,则 y =()A, cos(-x) - xs...