习题2.1 1、利用导数的定义求函数在定点的导数值。 (1)y=x3+1,x。=1(3) (2②)y=hx=e( 2、己知函数fx)=x3,求:f(x),f'(0),f"(x)。 (f"(x)=3x2,'0)=0,f'(xo)=3x2) 从我肥经苹频的加误发演接你对色是-巴飞+大子的职。 △r 4、请简述导数的几何意义,并谈谈你的理解。 5、求曲线y=x2上过点(1,2)处的切线方程,并确定哪一点的切线具有如下性质:(1)平 行于Ox轴:(2)与Ox轴成45°角。 习题2.1答案 1、(1)解△y=(x+△x)3+1-(x2+)=3x2△x+3x△x2+△x3 r+3xA+Ar=3+3xAx+Ar Ar m1=im(3×2+3×1×Ar+Ar2)=3 (2)Ay=H(x+Ar)-hx=(+) 1+ △x A2=k1+=+当号 -m+当品=m0+时-日 2、解△y=(x+△x)3-x3=3x2△x+3xx2+△x3 +Ar=+3xA+Ar Ar Ax f=(6r2+3xAr+△r)=3r2
1 习题 2.1 1、 利用导数的定义求函数在定点的导数值。 (1) 1, 0 1 3 y = x + x = (3) (2) y = x x = e 0 ln , ( e 1 ) 2、已知函数 3 f (x) = x ,求: f (x) , f (0), ( ) 0 f x 。 ( 2 f (x) = 3x , f (0) = 0 , 2 0 3 0 f (x ) = x ) 3、从我们已经掌握的知识出发谈谈你对 x f x x f x x y x x + − = → → ( ) ( ) lim lim 0 0 0 0 式子的理解。 4、请简述导数的几何意义,并谈谈你的理解。 5、求曲线 2 y = x 上过点(1,2)处的切线方程,并确定哪一点的切线具有如下性质:(1)平 行于 Ox 轴;(2)与 Ox 轴成 0 45 角。 习题 2.1 答案 1、(1)解 3 3 2 2 3 y = (x + x) +1− (x +1) = 3x x + 3xx + x 2 2 2 2 3 3 3 3 3 x x x x x x x x x x x y = + + + + = lim (3 1 3 1 ) 3 2 2 0 1 = + + = → = y x x x x (2)解 ln( ) ln ln ln(1 ) x x x x x y x x x = + + = + − = x x x x x x x x x x x x y 1 1 ln(1 ) ln(1 ) ln(1 ) = + = + + = x e x x x y x x x x x x x x x e 1 lim ln(1 ) ln[ lim (1 ) ] 1 0 1 0 = = + = + → → = 2、解 3 3 2 2 3 y = (x + x) − x = 3x x + 3xx + x 2 2 2 2 3 3 3 3 3 x x x x x x x x x x x y = + + + + = 2 2 2 0 f (x) lim (3x 3x x x ) 3x x = + + = →
所以f(0)=0,f(x)=3x2 3、4、(略) 5、解由例3可知y-2x,所以过点(1,2)的切线斜率为k=2 因此过该点的切线方程为y-2=2(x-)即y=2x (1)若切线的与Ox轴平行,则切线的斜率为0,即y=2x=0 所以过点(0,0) (2)由题意y=2x=tan45°=1 解得x分号 所以过点(兮宁的切线与0:输成45角 习题2.2 L.求下列函数的导数, (1)y=2WF-3x2+x ②y=-G+ (3)y=3+x3+33 wy=G+左) (5)y=x2e (6)y=e'Ix (7)y=2*cosxbg,x o)= (9)y=3x5 Inxsin x 0)y1+安 (1l)y=e*cosx-x3na+π2 (12)y=x2arctanx 2.求下列函数在给定点的导数值 (D已知f)=sin c,求f"(爱 @已蜘=写2+芳求@)
2 所以 2 0 3 0 f (0) = 0, f (x ) = x 3、4、(略) 5、解 由例 3 可知 y = 2x ,所以过点(1,2)的切线斜率为 k = 2 因此 过该点的切线方程为 y − 2 = 2(x −1) 即 y = 2x (1)若切线的与 Ox 轴平行,则切线的斜率为 0,即 y = 2x = 0 所以 过点 (0,0) (2) 由题意 2 tan 45 1 0 y = x = = 解得 2 1 x = , 4 1 y = 所以过点 ) 4 1 , 2 1 ( 的切线与 Ox 轴成 0 45 角 习题 2.2 1. 求下列函数的导数. (1) y = x − x + x 3 2 2 3 (2) 3 5 1 x x x y − + = (3) 3 3 y = 3 + x + 3 x (4) 1) 1 = ( +1)( − x y x (5) x y x e 2 = (6) y e x x = ln (7) y x x x 2 = 2 cos log (8) 2 sin x x y = (9) y 3x ln xsin x 5 = (10) 2 1 x x y + = (11) 3 2 y = e cos x − x ln a + x (12) y x arctan x 2 = 2. 求下列函数在给定点的导数值. (1) 已知 f (x) = sin x cos x,求 ) 6 ( f . (2) 已知 5 5 3 ( ) 2 x x f x + − = ,求 f (2)
)已知f)=F 1+ 求∫"(4) 3.求下列函数的导数。 ()y=sm5x+1) 2)y=1-x2)0o (3)y=cosx+e-* (④)y=-smx (⑤)y=hnx (6)y=arcsin x2 (7)y=2sin +tan2x 8)y=V1+x3 (0)y-In tan (10)y=sin cosx3 (11)y=Mx+V1+x2) (12)y=sin1+x 画号 y=e2* (15)y=(arcsin x)2 (16)y=(sin√F) 4求下列方程所确定的隐活数的号数安 (1)x3+y3-3xy=0 (②)xy=e" (3)y=cos(x+y) (4)sin(xy)=x 6求下列参发方程所确定高数的导数会 (1) x=1-12 (2) [x=e'sint y=t-13 ly=e'cost 3at (3) x1+ (4) [x=n1+t2) y=t-arctan y-1+7 7.利用对数求导法求下列函数的导数, (1)y=xsnx (2)y= √x+2(3-x)1 (r+) 8.求下列函数的高阶导数 (1)已知y=4x2+hx,求y. 3
3 (3) 已知 x x f x + − = 1 1 ( ) , 求 f (4) . 3. 求下列函数的导数. (1) y = sin( 5x +1) (2) 2 100 y = (1− x ) (3) x y x e − = cos + (4) 3 y = 1− sin x (5) y = ln ln x (6) 2 y = arcsin x (7) y x x 2 tan 2 sin = + (8) 3 y = 1+ x (9) x y 1 = ln tan (10) 3 y = sin cos x (11) ln( 1 ) 2 y = x + + x (12) 2 y = sin 1+ x (13) 2 tan x y = (14) x x y e + = 2 (15) 2 y = (arcsin x) (16) 3 y = (sin x) 4.求下列方程所确定的隐函数的导数 dx dy . (1) 3 0 3 3 x + y − xy = (2) xy xy = e (3) y = cos(x + y) (4) sin( xy) = x 6.求下列参数方程所确定函数的导数 dx dy . (1) = − = − 3 2 1 y t t x t (2) = = y e t x e t t t cos sin (3) + = + = 2 2 2 1 3 1 3 t at y t at x (4) = − = + y t t x t arctan ln(1 ) 2 7. 利用对数求导法求下列函数的导数. (1) x y x sin = (2) 5 4 ( 1) 2(3 ) + + − = x x x y 8. 求下列函数的高阶导数. (1) 已知 y 4x ln x 2 = + ,求 y
(2)已知y=x3hx,求y (3)己知f(x)=(x+10)6,求f"(2). (4)已知y=n(1-x2),求y” 9.求下列函数的n阶导数. (1)y=e2 (②)y=(1+x) 习题2.2答案 1.求下列函数的导数. (1)y=2F-3r2+x 2)y=产-F+1 解y=(2F-3到2+x 解y=(x2-x号+x =(2-(3Fy+(xy =(x2y-(x)y+(x3) 左2安1 2x*2 (3)y=3+x3+3 ④=+左- 解y=(3+x3+3y 解-G+存旷 =(3y+(x3y+(33y + =3n3+3x2 =r*( 11 =22 (5)y=x2e (6)y=e'Inx 解y-(x2ey 解y'-(ehx =(x2)'e*+x2(e")' =(e*)'In x+e"(In x)
4 (2) 已知 y x ln x 3 = ,求 y . (3) 已知 6 f (x) = (x +10) ,求 f (2) . (4) 已知 ln(1 ) 2 y = − x ,求 y . 9. 求下列函数的 n 阶导数. (1) x y e 2 = (2) y = ln(1+ x) 习题 2.2 答案 1.求下列函数的导数. (1) y = x − x + x 3 2 2 3 (2) 3 5 1 x x x y − + = 解 (2 3 ) 3 2 y = x − x + x 解 ( ) 2 2 3 5 = − + − − y x x x =(2 ) (3 ) ( ) 3 2 x − x + x ( ) ( ) ( ) 2 2 3 5 = − + − − x x x 1 1 2 1 3 = − + x x 3 4 3 2 5 2 x x x = x + − (3) 3 3 y = 3 + x + 3 x (4) 1) 1 = ( +1)( − x y x 解 (3 3 ) 3 3 y = + x + x 解 1) 1 = (1− + − x y x (3 ) ( ) (3 ) 3 3 = + x + x ) 1 = (− + x x 2 3 ln 3 3x x = + ) 1 = (− ) + ( x x x 2x x 1 2 1 = − − (5) x y x e 2 = (6) y e x x = ln 解 ( ) 2 = x y x e 解 y = (e ln x) x =( ) ( ) 2 2 + x x x e x e =(e )ln x + e (ln x) x x
=2xe'+x2e' =e'hxte'l =xe'(2+x) -e(hx+) (7)y=2*cosxlg:x 解y=(2cosx10g2x) =(2)'cosx lgx+2*(cosx)'logx+2*cosx(og2 x)' =2rn2cosxbg2x-2snxbg2x+2cosxxh2 解y=0 _smx水x2-snx-2y -xcosx-2xsin x (9)y=3x'In xsin x 解y'=(3x3 n xsin x)' =(3x5)'In xsin x+3x5 (In x)'sin x+3x5 In x(sin x)' =3x(5In xsin x+sin x+xInxcosx) (10)y=1+x (11)y=e'cosx-x'na+ 解广= 解y'=(e2cosx)'-(x3nay+(π2)y -0+)-x0+x2y =e*cosx-e*sin x-3x2 In a (1+x2)月 (12) y=x2 arctanx 解y'=(x2 arctanx)
5 = x x xe x e 2 2 + x e x e x x 1 = ln + xe (2 x) x = + ) 1 (ln x e x x = + (7) y x x x 2 = 2 cos log 解 (2 cos log ) 2 y = x x x (2 ) cos log 2 (cos ) log 2 cos (log ) 2 2 2 = x x + x x + x x x x x ln 2 1 2 ln 2cos log 2 sin log 2 cos 2 2 x x x x x x x x x = − + (8) 2 sin x x y = 解 ) sin ( 2 = x x y 4 2 2 (sin ) sin ( ) x x x − x x = 4 2 cos 2 sin x x x − x x = (9) y 3x ln xsin x 5 = 解 (3 ln sin ) 5 y = x x x (3 ) ln sin 3 (ln ) sin 3 ln (sin ) 5 5 5 = x x x + x x x + x x x 3 (5ln sin sin ln cos ) 4 = x x x + x + x x x (10) 2 1 x x y + = (11) 3 2 y = e cos x − x ln a + x 解 ) 1 ( 2 + = x x y 解 y ( cos ) ( ln ) ( ) 3 2 = e x − x a + x 2 2 2 2 (1 ) ( ) (1 ) (1 ) x x x x x + + − + = e x e x x a x x cos sin 3 ln 2 = − − 2 2 2 (1 ) 1 x x + − = (12) y x arctan x 2 = 解 ( arctan ) 2 y = x x
=(x2)'arctanx+x2(arctanx)' 2 =2xarctan 2.求下列函数在给定点的导数值, 1)已知f)=snxc0sx,求f"爱 解f'(x)=((sin xcosx) =(sin x)'cosx+sin x(cosx)' (cosx)2-(sin x)2 =cos2x fr月=os2=os号-月 解=写2号r -写25 62可号 o-号吕 @)已知f=1- 1+F 求∫"(4) 四-片授 _0-0+-0-FX1+ +2 =x+网 6
6 ( ) arctan (arctan ) 2 2 = x x + x x 2 2 1 2 arctan x x x x + = + 2. 求下列函数在给定点的导数值. (1) 已知 f (x) = sin x cos x,求 ) 6 ( f . 解 f (x) = (sin x cos x) = (sin x)cos x + sin x(cos x) 2 2 = (cos x) − (sin x) = cos2x 2 1 3 cos 6 ) cos 2 6 ( = = = f (2) 已知 5 5 3 ( ) 2 x x f x + − = ,求 f (2) 解 ) 5 5 3 ( ) ( 2 + − = x x f x ) 5 ) ( 5 3 ( 2 + − = x x x x 5 2 (5 ) 3 2 + − = 15 17 5 4 3 1 f (2) = + = (3) 已知 x x f x + − = 1 1 ( ) , 求 f (4) . 解 ) 1 1 ( ) ( + − = x x f x 2 (1 ) (1 ) (1 ) (1 )(1 ) x x x x x + − + − − + = 2 (1 ) 1 x + x = −
f0=-i8 1 3.求下列函数的导数 (1)y=sim5x+1) 解y'=sim(5x+1)=cos(5x+1)-(5x+1)y=5cos(5x+1) (2)y=0-x2)1 解y=[1-x2)100y=1001-x2)9.1-x2y=-200.x1-x2)9 (3)y=cosx+e-x y'=(cosx+e)'=(cos)'+(e-)'=-sin ('+e-".(-x)' 2-e (④)y=-snx 解y=(-snxy=0-smxj=1-sm)手.1-snxy COSX 0-m对 (⑤)y=nnx 1 (6)y=arcsin x2 解y=amry-京y-产 (7)y=2sins +tan2x y'=(2tim +tan2x)'=(2sin)+(tan2x)' =2snx In 2.(sin x)'+sec22x.(2x) =2n cosxIn 2+2sec22x (⑧)y=V+x 7
7 18 1 f (4) = − 3. 求下列函数的导数. (1) y = sin( 5x +1) 解 y = [sin( 5x +1)] = cos(5x +1)(5x +1) = 5cos(5x +1) (2) 2 100 y = (1− x ) 解 [(1 ) ] 100(1 ) (1 ) 2 100 2 99 2 y = − x = − x − x 2 99 = −200x(1− x ) (3) x y x e − = cos + 解 = (cos + ) −x y x e = (cos ) + ( ) −x x e = −sin ( ) + (− ) − x x e x x x x e x − = − sin − 2 1 (4) 3 y = 1− sin x 解 ( 1 sin ) [(1 sin ) ] 3 1 3 y = − x = − x (1 sin ) (1 sin ) 3 1 3 2 = − − − x x 3 2 3 (1 sin ) cos x x − = − (5) y = ln ln x 解 (ln ) ln 1 = (ln ln ) = x x y x x ln x 1 = (6) 2 y = arcsin x 解 ( ) 1 1 (arcsin ) 2 4 2 − = = x x y x 4 1 2 x x − = (7) y x x 2 tan 2 sin = + 解 (2 tan 2 ) (2 ) (tan 2 ) sin sin y = + x = + x x x 2 ln 2 (sin ) sec 2 (2 ) sin 2 = x + x x x x x x 2 cos ln 2 2sec 2 sin 2 = + (8) 3 y = 1+ x
解y=(+y=I+x)=+x).0+xy oy=h时 解y=an=m=- 产an (10)y=sin cosx' y'=coscosx3 .(cosx')'=coscosx3.(-sin x)-(x3)' =-3xcoscosx'sinx (11)y=Mx+1+x2) 解y'=m(x+V1+x2川= ++++平y +i+字+Mn=1 1 ++0+ (12)y=sin+x y'=(sin 1+x2)'=cos+x2(+x2)' :s+7+ =xcos1+x V1+x2 )= 解y=(amy=an=an2(an 8
8 解 (1 ) (1 ) 2 1 ( 1 ) [(1 ) ] 2 3 1 2 3 1 3 3 = + = + = + + − y x x x x 3 2 2 1 3 x x + = (9) x y 1 = ln tan 解 ) 1 (tan 1 tan 1 ) 1 = (ln tan = x x x y x x x 1 tan 1 sec 2 2 = − (10) 3 y = sin cos x 解 cos cos (cos ) 3 3 y = x x cos cos ( sin ) ( ) 3 3 3 = x − x x 2 3 3 = −3x cos cos x sin x (11) ln( 1 ) 2 y = x + + x 解 ( 1 ) 1 1 [ln( 1 )] 2 2 2 + + + + = + + = x x x x y x x [( ) ( 1 ) ] 1 1 2 2 + + + + = x x x x ) 1 (1 1 1 2 2 x x x x + + + + = 2 1 1 + x = (12) 2 y = sin 1+ x 解 (sin 1 ) cos 1 ( 1 ) 2 2 2 y = + x = + x + x 2 2 1 cos 1 x x x + = + 2 2 1 cos 1 x x x + + = (13) 2 tan x y = 解 ) 2 ) (tan 2 (tan 2 1 ) ] 2 ) [(tan 2 ( tan 2 1 2 1 = = = x x x − x y
4/tan (14)y=e 解y'=(e*r)=ea.(x2+x)=e+r(2x+1) (15)y=(arcsin x) y'=[(arcsin x)2]=2arcsin x(arcsin x)' =2arcsinx V1-x2 (16)y=(sin x) 解y'=[sim√)r=3sin√F)2(sin x)'=3(smV)2,cosF.(F)y 3s)'cos 2vx 4求下列方君所南定的隐西激的导数产 (1)x3+y3-3xy=0 解方程两边分别对x求导数,得 (x3+y3-3x)x=0 (x3),+0y)-(3x,=0 3x2+3y2y-30y+y)=0 30y2-x)y=3y-x2) ” x-y (②)xy=e 解方程两边分别对x求导数,得 9
9 2 4 tan 2 sec 2 x x = (14) x x y e + = 2 解 ( ) ( ) 2 2 2 = = + + + y e e x x x x x x (2 1) 2 = + + e x x x (15) 2 y = (arcsin x) 解 [(arcsin ) ] 2arcsin (arcsin ) 2 y = x = x x 2 1 2arcsin x x − = (16) 3 y = (sin x) 解 [(sin ) ] 3(sin ) (sin ) 3(sin ) cos ( ) 3 2 2 y = x = x x = x x x x x x 2 3(sin ) cos 2 = 4.求下列方程所确定的隐函数的导数 dx dy . (1) 3 0 3 3 x + y − xy = 解 方程两边分别对 x 求导数,得 ( 3 ) 0 3 3 = + − x x y xy ( ) ( ) (3 ) 0 3 3 = − + x x x x y xy 3 3 3( ) 0 2 2 x + y y − y + xy = 3( ) 3( ) 2 2 y − x y = y − x 2 2 x y x y y − − = (2) xy xy = e 解 方程两边分别对 x 求导数,得
(xy):=(e). y+xy=e(y+xy) x(1-e)y'=ye-1) 片 (3)y=cos(x+y) 解方程两边分别对x求导数,得 y:=[cos(x+y) y'=-sin(x+y)(I+y) sin(x+y) y=1+sx+列 (④)si(xy=x 解方程两边分别对x求导数,得 [sin(xy)=1 cos(xy)(Uy+y')=1 y'=1-ycosxy xcosxy 5求下列参数方程所确定函数的导数 x )r=1-2 y=1-2 解=-1-3- 0-2y -21 21 (2)(x=e'snt y=e'cost 会名m cost+sin 6
10 x xy x (xy) = (e ) y xy e (y xy ) xy + = + (1− ) = ( −1) xy xy x e y y e x y y = − (3) y = cos(x + y) 解 方程两边分别对 x 求导数,得 x x y = [cos(x + y)] y = −sin( x + y)(1+ y ) 1 sin( ) sin( ) x y x y y + + + = − (4) sin( xy) = x 解 方程两边分别对 x 求导数,得 [sin( xy)] x =1 cos(xy)( y + xy ) = 1 x xy y xy y cos 1− cos = 5.求下列参数方程所确定函数的导数 dx dy . (1) = − = − 3 2 1 y t t x t 解 t t t t t dx dy 2 1 3 (1 ) ( ) 2 2 3 − − = − − = t t 2 3 1 2 − = (2) = = y e t x e t t t cos sin 解 ( sin ) ( cos ) = e t e t dx dy t t t t t t cos sin cos sin + − =