極限limx3 xx6

(e3/2函數(shù)(1x2arctanx的二階導數(shù)是2arctanx

。14.limxx 注：x xxxx x注：x xxxx x x2x

極限

1 x0(sin x2 x2(sin x x2(sin注：由于(sin

x2

x2(sin sinx x

x

x1

2o(x3

1x

o(x4) 故 11 x0(sin x2

x2函數(shù)yarctanxex的反函數(shù)導數(shù)為 ex(x21)若可微函數(shù)y 由參數(shù)方程xtet，ytf

2(te2t 11極限

2

1f(x)

112

x0（或第一類間斷2sin2xe2ax若函數(shù)f(x)

xx

x0處連續(xù)，則a2

sin

xsin(2x1

sin(2x1)yx

)（x0的微分dy

)2cos(2x1)lnx

dx 若當x0時，函數(shù)ln(1x1x)與函數(shù)xp為等價無窮小，則p 的n階導數(shù)為n

。1x 2(1 (1x)n1x-yxt2sint,ytcost于點(0,1處（即t0)的切線方程為yx1。

1x1處的帶Peano余項的nTaylorx1 1(x

1x1x1)2(1)n(x1)no(x1)n

1limsinxlimsinx sinx1cos cosxlim

limln|sinx|ln|x|=limsin x

1cos sin=limxcosxsin = xsin = xsin2 x0sin2x2xsinxcos

sinx1cosx1sinxxsinx

1cosx1x 1xsinx而

1x3o(x3

x0 1cos x[1x2o(x2 1/sinx1cosx因此

。解答完畢。假設由方程x3y3xy10確定了一個在x0解：于方程x3y3xy10x0，得y310。因此y(0)1。對方程x3y3xy10兩邊關于x求導得3x23y2yyxy0x0y1得3y10y(013。6x6y(y)23y2y2yxy于上式中令x0y1y13得3y0y(0)0。Maclaurin展式為y(x)11xo(x2。解答完畢。3f(x)|x(x21)|在閉區(qū)間[0,2]f(xf(xf(x在開區(qū)間(0,2)上僅有一個x1f(xsgn(x(x21))](3x21)x0,1x1。故f(x)在開區(qū)間(0,2)內有且僅有一個駐點1/3。 3Mmaxf0f1,f(1),f(2)max{0 0,6} 3f(x僅在閉區(qū)間[0,2]x2處取得它的最大值6yx2上一個點(xyy1上一個點(xy 1在點(x1y1yx在點(x2y2處的切線相同。并寫出這條公共切線的1yx2在點(xyxx2 yx22x(xx)或y2xx y1在點

1 2,y2)(x2,2 y 2(xx2)

y x x 假設點(x1y1和點(x2,y2)滿足題目的要求，則由式（*）和（**）所定義的直線相同。xx應滿足方程組2x1x22

x (xx(2,1。于是所求的兩個點分別為(xy2,4和(xy122 y4x4證明方程x52x310有且僅有一個正根f(xx52x31。則f(0)1f()。因此方程x52x3165至少有一個正零點。由f(x)5x46x2x2(5x26)可知，函數(shù)f(x)有唯一的正臨界點 65f(x)0x(0,。故f(x在區(qū)間(0,f(x)0x(,f(x在區(qū)間(,f(f(x)f(0)1x(0,。由此可見函數(shù)f(x有且僅有一個正零點，并且這個零點位于開區(qū)間(,。證畢。設f(x在[0,1上連(0,1上可導，且滿足|f(x|1x(0,1)f(0f(1。證明|f(x1f(x2|12x1x2(0,1證明：當|x1x2|1/2Lagrange|f(x1f(x2||f(||x1x2|12。結論成當|x1x2|12時，不妨設0x1x21x2x112。于|f(x1)f(x2)||f(x1)f(0)f(