33.25考研数学基础计算150题-求导答案【公众号：小盆学长】免费分享.pdfVIP免费

求导计算题及答案0002221.()sin,lim().()sincos,222222lim2lim2222hhhfhffxxxhfxxxxfhffhffhh设则解析：由已知可得故220002102.()(e1)(e2)(e),(0)().()(0)(e1)(e2)(e)(0)limlim0lim(e2)(e.)(12)(1)(1)(1)!xxnxxxnxxxxxnxnxfxnnffxfnfxxnnnZ设则解析：由得，,导数定义22.3(),(ln),().(ln)(ln)(ln),1(ln)(ln)(ln)(ln).yfxyfxyfxyfxxxfxxfxfxfxxyxx设函数二阶可导若则解析：由于故00000000(3)(2)4.(0)2,lim.sin(3)(2)(3)(2)limlimsin(3)(0)(2)(0)(3)(0)(2)(0)limlimlim(3)(0)(2)(0)3lim2lim3(0)232xxxxxxxxfxfxfxfxfxfxfxxxfxffxffxffxfxxxfxffxffxx设则解析：(0)5(0)5210.ffsinsinsinlnsinlnsin()(0),()________.()e,sin()esinlncosln.5.xxxxxxxfxxxfxfxxxfxxxxxxx设则解析：由于故1()()()()()(1)()1()126.()ln,(0)________.1312()lnln(12)ln(13),13()[ln(12)][ln(13)],ln(1)(1)(1)!(1),(1)1![(2)12313],(0nnnnnnnnnnnnnnnxfxfxxfxxxxfxxxaxanaxfxnxxf设则解析：因为所以又故1)(1)(1)![(2)3].nnnn2222222321ln1d,2darctandd/d1,dd/ddddddd111.dddddd1xtyyyxxytyytxxttyyytttxxxtxxttt7.设由参数方程确定则解析：由于故222232d8.=()sin()ddddcos()1cos()1,1,ddd1cos()1cos()dsin()x(1).d1cos()1cos()yyyxyxyxyyyxyxxyxxxxyxyyxyyyxxyxy设由方程确定，解析：方程两边对求导得解得上式两边再对求导得0000222ln109.:0,.1,00,ln(1)111()(0)ln(1)11(0)limlimlimlim;0221ln(1)(1)ln(1)10,(),(1)(,()xxxxxxfxfxxfxxxxxfxfxxxxfxxxxxxxxxxxfxxxxxfx设证明在...