2-1高等数学同济大学第六版本

习题211[0t]从而转角t的函数那么称如果t0旋转是非匀速的应怎样确定该物体在时刻t的角速度？0解在时间间隔[ttt]内的平均角速度为0 0 (tt)(t)t 0 t 0 故t0时刻的角速度为

(t

t)(t)lili

lim

0(t)t0 t0t t0 t 02当物体的温度高于周围介质的温度时物体就不断冷却若物体的温度0 TtTT(t)t解物体在时间间隔[ttt]0 TT(tt)T(t)平均冷却速度为T(tt)T(t)故物体在时刻t的冷却速度为limlimT(tt)T(t)T(t)t0t0 3xf(x)元f(x)称为成本函f(x)f(x)在经济学中称为边际成本f(x)的实际意义解f(xx)f(x)表示当产量由x改变到xx时成本的改变量f(xx)f(x)表示当产量由x改变到xx时单位产量的成本xf

f(xx)f

表示当产量为x时单位产量的成本x0 4f(x)10x2f(1)解f(1)lim

f(1x)f

10(1x)210(1)2x0 x x0 x10limx0

2xx2x

10lim(2x)20x066下列各题中均假定f(x)指出A表示0什么(1)limf(xx)f(x)A0x0x0解Alimf(xx)f(x)0x0limx0f(xx)f(x)f0x0x0(x)0(2)limf(xAf(0)0f(0)存在x0x解Alimf(x)limf(0x)ffx0xx0x(3)limf(xh)f(xh)A0h0h0解Alimf(xh)f(xh)0h0h0lim[f(xh)f(x)][f(xh)f(x)]0000h0hlimf(xh)f(x)limf(xh)f(x)0h0h000h0hf(x)[f(x)]2f(x)0007求下列函数的导数(1)yx4(2)y3x2(3)yx165证明(cosx)sinx解(cosx)limcos(xx)cosxx0xlimx02six)si22xlim[si2) 2]sixsinxx0x2(4)(4)y1 x(5)y1x2(6)yx35x(7)yx(7)yx23x2x5解(1)y(x4)4x414x3(2)y(3x)(x) x x223221233313(3)y(x16)16x16116x06(4)y(1x)(x ) x121111 22 x232(5)y(1)(x2)2x3x2(6)y(x1635x)(x ) x51651651 x 161155(7)y(x x2 x53)(x) x x 1611115666688st3(m)t2秒(s)时的速度v(s)3t2v|t212(米/秒)9f(x)为偶函数f(0)存在f(0)0证明当f(x)为偶函数时f(x)f(x)所以fli(x)fli(x)flimf(x)ffx0x0x0x0x0x02f(0)0f(0)010求曲线ysinx在具有下列横坐标的各点处切线的斜率 x2x3解因为解因为ycosx所以斜率分别为kkcos1132kcos1211求曲线ycosx上点(,1)处的切线方程和法线方程式3 2解ysinyxsin3332故在点(,1)切线方程为y1 3(x)3 2 2 2 3y12(x2 3 312yex在点(01)处的切线方程解yexy|x

1故在(01)处的切线方程为0y11(x0)即yx101 13yx2x1x31 解y2x割线斜率为k2x4x2

91431 2yx2上点(24)处的切线平行于这条割线14x0处的连续性与可导性(1)y|sinx|x2sin1 x0(2)y 0解(1)因为

x0y(0)0 limx0

ylimx0

|sinxlim(sin0x0limlimylim|sinxlimsinx0x0 x0 x0x0处连续又因为mlimy(x)lim

|six||si0|lisix1

x0

x0

x0

x0 xy(0)y(0)limy(x)lim|sinx||sin0|limsi1x0x0x0x0x0xx0处不可导解因为limy(x)limx2sin10y(0)0x0处连续x又因为limx0y(x)x0lix0x2si0x1xliin0x0xx0处可导y(0)01515设函数f(x)bx2xx1f(x)x1处连续且可导ab应取什么值？解因为lim

f(x)limx21

f(x)lim(axb)abf(1)abx1

x1

x1x1处连续ab1又因为当ab1时fli2xf(1)limaxb1lima(x1)ab1lima(x1)ax1x1x1x1处可导a2b116已知f(x)x2xx0x0f(0)f(0)f(0)是否存在？解因为解因为f(0)limf(x)f(0)limx01x0xx0xf(0)limf(x)flimx200x0xx0xf(0)f(0)f(0)不存在1717f(x)xxx0x0f(xx<0时x<0时f(x)sinf(x)cosx当x>0时f(x)xf(x)1f(0)limf(x)flimsinx01xf(0)limf(x)flimx01f(0)1从而x0x0xx0xx0xf(x)x1x0x018证明双曲线xya2上任一点处的切线与两坐标轴构成的三角形的面积都等于2a2解由xya2得ya2 kya2xx2设(x0y0)为曲线上任一点