第三章习题3.2答案.docx
1.推导余切函数及余割函数的导数公式:解:(cot %)=cosx'、sin x ,(cosx) sinx-cosx(sinx)sin2 x 22-sin x-cos xsin2 x= -csc2 X2.求下列函数的导数:7?(1)广人 + 12x x解：y=3f 芝+义.(2) y = 5V2" + 3/解：y=l5x2-2x ln2 + 3e(3) y = 2tanx + secx l解：y = 2 sec2 x + sec x tan x.(4) y = sinxcosx解：y =cos2 x-sin2 x = cos2x(5) y = Y inxI解：y = 2xnx + x.(6) y = 3, cosx解：y = 3ex cos x - 3ex sin x = 3eA (cos x - sin x).(7)Inx尸一x解:、1 - In x y = 厂ex(8) ,y = + In3exx2 -2xex _ (x-2)evx4x3(9) y = x2(lnx)cosx解：y = 2%(ln x)cos x+xcos xx1 (in %) sin x.(1 + X221 -%1 + X10.设函数/(x)和g(x)可导，且r(x) +f(x) w 0.试求函数y = J/2 (x) + g2 (无)的导 数。解：，_/(x)/(x) + g(x)§(x) J(%) + g2(%)1L设/(x)可导，求下列函数的导数:(2) y = / (sin2x + f (cos2x)解：12 .求下列函数的导数：(1) y = ex 一 2%+3)解：y = ex解：y = exx2-2x+3)+ e-x(2x-2)= ex (-x2+4x-5)(2) y = sin2 %-sin产解：y =sin2xsin(%2)解：y =sin2xsin(%2)+ 2xsin2 xcosl x2/、2X arctan 12；彳 x4 arctan 解：y' = 2 arctan =31 . x2 4 + x21 + 4/ 、 Inx(4) y =xnt 一nxn Inx 1-nlnxx2nd - er k二-sin I 41 tan-x x2y - Ineos x解：y = cos X-sin1 2 y-e x-2sin-cos- 2 -sin2- sin e xxx2解：y=J2x + x(9) y = xarcsin + a/4-x2 21解：y4-x213 .求下列函数的微分:X. X,=arcsin (5)y =Inx&刀 .4xd In x - In xd yx 2-lnx .解：dy = axx2xz(6) y =解：dy = f=dx +a2(7)y = tan2%+n cosx解：dy =(2 tan xsec2 a:- tan x6Zr =(2tan y = tanxH! cosx x+tan xdxr2 4(8) y = e cos x解：14用微分的运算法则求下列函数的微分:(1)y = x2 +4x + l解:(2)x2-l解：力=x3一 1) (Y ”(工3 +)一j +3f +2工dx解：dy = d tan x +cos xdl - d cos x 1 + sin xCOS2 Xcos2 Xdx2(4) y = cosx”解：dy =-sin x2d) = -2xsin jdx(5)1x解:-/ dx(6) = arctan (in x).7J7 dlnx1dx解：dy ='+ ln x xl + n x15 .设卬(九)都是x的可微函数，求下列函数的微分:(1) y = u-v-w解：dy = vwdu + uwdv + uvdw(2) y = Inyju2 +v2f 1/ 9解：dy-d -Inr + vudu + vdv22u(3) y = arctan v(a 解：dy = fu+7vdu - udv22U +V3-解：dy = ir + F +w2 + v2 + w2) = 3 (2 + v2 + w2 )5 (udu + vdv + Wvv)(5) y = e"解：dy = ellvd (wv) = eltv (ydu + udv)(6) y = ev sin u解：dy = sin udev + evd sin u - sin uevdv + cos uevdu(7)arctan (wv)arctan(wv)arctan解：dy = 6aarctan(wv) = TTd() = -y(vdu + udv)16 .设函数x)和g(x)均在点与的某一邻域内有定义，/(x)在/处可导，%) = 0,g(X)在4处连续，试讨论f (x) g(X)在4处的可导性。解：所以/(x)g(x)在与处可导。17 .设函数/(x)满足下列条件：(1) /(%+y) = /(%)/3)，对一切,yeR;(2) x) = l + xg(x),而limg(x) = l.试证明在R上处处可导，且/(x) = x).f(o) = l ，所以证明：在 x) = l + xg(x) 中令 x = 0f (O) =lim/G)T= limg(x) = 1。' 'x->0 x 0x->0%x->0'，/ 、1 + sinf(10) s =+ cos t解：scos r(l + cos，) + sin / (1 + sin1 + cos t + sint(1 + cos/3.求下列函数的微分：(1) y = + 2>/xx)2(1 + cosZ)2解：dy =dx.1 u +(2) y = xsin2x(3) y = xsin2x解：dy = sin 2xdx+az/(sin 2x) = (sin lx+2xcos 2x) dx.z 、X(4) y = = Jf+i解：dy =解：dy = z-： n x2 + Idx -7 x +ldx-xdx +1x2J/ +1dxdxx2+lx2+lx2 +(4) = ln2(l-x).21n(l-x)解：dy =-dx.x-解：力=e2xd (x2) + xW=2(x + x2) e2xcbc.(6) y = ”'cos(3-x)解：dy = cos(3 - x) dex + exd cos(3 - %) = (sin(3 - x) - cos(3-x)exdx.(7) y = arcsinJl-V5 1 1 解：dy = j= 7Hl.-x2(8) y = tan2(l + 2光之)Vl-x7解：dy = Sx tan (1 + 2x2) sec2 (1 + 2/)小：.(9) y = arctan匕1 + x2解：dy =i - x2 Yl + x2>dx = dx.1+/(10) s = Asin3 + °).解：ds = coAcoscot + cp)dt.4.以初速度%竖直上抛的物体,其上升的高度S与时间/的关系是s = %，（g/.求:（1）该物体的速度u（。；（2）该物体达到最高点的时刻。解：(1) y(O = s (/) = %-go（2）该物体达到最高点的时刻速度应为0,所以,二员. g5.求曲线 =2sinx + V上横坐标为 = 0的点处的切线方程和法线方程。解：y（O）=（2cosx + 2x）1_0 =2. y（0）= 0,所以切线为 y = 2x,法线为 y = ，x.1八y* Q| n Y I I6 .讨论函数力= F，在x = 0处的连续性与可导性。0,x = 0.解：lim/(x) = limxsin = 0 = /(0),所 以 /(x)在 x = 0 处连续 理坦锣=!吧拈不存在,所以小）在=。处不可导。7 ,求下列函数的导数: （1） y =（2x + 5）4解:y =8(2x+5)(2) y = cos(4 3x)解：y =3sin(4-3x).解:y = -6xe3x(4) y = ln(l + %2)(5) = sin2x解：y = 2 sin x cos x = sin 2x(6) y a1 x1(7) y = tanx2解：y = 2xsec2 x2.(8) y = arctan(，)(9)(9)y = (arcsinx)22arcsin x(10) y = ncosx-sinx=一 tan xcosx/ 、 1 2 1 3y x x Hx 23解：y =l-x + x2.,、 1 1 1(12)五+双.11 1解：厂一丁力7一七七z 、 ax + b(13) y =cx + dcx + dy解 y _ a(cx + d)-c(ajc + b) _ ad-becx + d)2(14) y = (x-6z)(x-/?)2(x-c)3)3 +3(X-6Z)(X-Z?)2 (x-c)2 .)3 +3(X-6Z)(X-Z?)2 (x-c)2 .(15). sinx y = xsmx-解：y=sinx + xcosx +xcosx-sin jcx2(16)y = x10"解：y =(x-/?)2 (x-c)3 4-2(x-6z)(x-Z?)(x-c= 10x +xA0xn0.8.求下列函数的导数:(1) y = arcsin (1- 2x)解：y=."241-(1-2x)2-x1-x13(-2力=-J(3) y = e 2 cos3x-e 2 cos3x-3e 2 sin 3x1xr7解：y = 一(4)= arcsin-1yjx2 -1/ 、 1-lnx(5) y =1 + lnx解：y =?(l + lnx)x(l + lnx)/八 sin lx(6) y =xe . 2x cos 2x-sin 2x解：y =、x(7) y = arcsinyfxXX9 + JT解:X + J/ + %1 21la2 +x2(9) y = ln(secx+tanx)2sec x tan x + sec x =sec x.sec x + tan x(10) = In (escx-cot x)2-cscxcotx + csc X =CSC X.CSC x-cot X(11)y = arcsin sin xcosxCOS XVl-sin2 xCOS X(12)i+xY1 + x2(13), X,y = In tan cosx-In tanx 2解：(14) y = n(ex+yll + e2xexV1 + e2xexV1 + e2xex + 解:y=x . /i . 2xe +l + e_i_(15) y = x“x>0)(止、 处_n y |-2解：-e x = jc x (1-lnx).< ) x(16) y = eclx sinbx=aecix sin bx + beax cos bx(17)y = arcsin a(18)(18)sgn。y/a2 -x2y = arctan a11 aa 1 x2l + - a(19)(19)y = cos5 x=-5 cos4 xsinx(20) y = lntan3x.3sec-3x 6：y =tan 3x sin 6x/2(21) y = lnt=(I 2 t解：y - 2In t+=I2 ' j) t + t(22)y = arcsin2xx2+l2(x2+1)-4x2(x2 + 1)2sgn(l-x2)(2x )2 -1 + f、Y + 1 /解：y1 a-b 2%c -4 /see .22y1 a + b 2 _1J/ 义-、!？” ei + bcosxa + h2(24)y = n la解:(25) y = _ Jx? + er H In x + J尤？ + er (aw22、解：(26) y = dx? - /In x + dx- - er ( w 0)解：9 .求下列函数的导数：(工丫(1) y = arcsin I2Jc X解:解:(2)2arcsin -解:12 x一 sec -i221y - X tan- 2sinx(3)y = yjl + n2 x1 InxInx解:y- J+ ln2 x(4)_ arctan x/x y - e解:,'_ arctan xfxy =earctan x/x1 + x 2x 2yx (1 + x)(5)解:' n n 一1/ y =sin xcosxcosnx-nsn xsmnx = nsm xcos( + l)xarcsin xy 二arccosx解:arccos x arcsin x yjl-x1 J-/arccos x)2arcsin x + arccos xyji-x1 (arccos %)2(7) y = In In In xI解：= -J- In Inx xlnx(ln In x) 解:(9)x2Vl-x2-x2x21-V1-X2X2y/1-X21 + x y = arctan1-x2
