2024年2月21日发(作者：聊乐蓉)

高等数学第六版课后习题及答案 第二章第四节

习题 24

1 求函数的二阶导数

(1)

y2x2ln

x

(2)

ye2x1

(3)

yxcos

x

(4)

yet sin

t

(5)ya2x2

(6)

yln(1x2)

(7)

ytan

x

(8)y31

x1 (9)

y(1x2)arctan

x 

xe (10)y

x (11)yxex

(12)yln(x1x2)

解 (1)y4x1

y412

xx (2)

ye2x1

22e2x1

y2e2x1

24e2x1

(3)

yxcos

x



ycos

xxsin

x

ysin

xsin

xxcos

x2sin

xxcos

x 

(4)

yetsin

tetcos

tet(cos

tsin

t)

yet(cos

tsin

t)et(sin

tcos

t)2etcos

t 

2

(5)y1x

(a2x2)22222axaxa2x2xxa2x22

ya2xa2

2222(ax)ax (6)

y12(1x2)2x2

1x1x2(1x2)2x(2x)2(1x2)

y

(1x2)2(1x2)2 (7)

ysec2

x

y2sec

x(sec

x)2sec

xsec

xtan

x2sec2xtan

x 

2(x31)3x (8)y3232

(x1)(x1)6x(x31)23x22(x31)3x6x(2x31)

y

(x31)4(x31)3 (9)y2xarctanx(1x2)122xarctanx1

1x

y2arctanx2x2

1xxxex(x1)exe1 (10)y

x2x2[ex(x1)ex]x2ex(x1)2xex(x22x2)

y

x4x3 (11)yexxex(2x)ex(12x2)

yex2x(12x2)ex4x2xex(32x2)

(12)y11(x1x2)(12x)1

x1x2x1x221x21x2222222x

y12(1x2)122x

1x1x21x2)(1x)21x 2 设f(x)(x10)6

f (2)?

解f (x)6(x10)5

f (x)30(x10)4

f (x)120(x10)3

f (2)120(210)207360

d2y 3 若f (x)存在 求下列函数y的二阶导数2

dx3 (1)

yf(x2)

(2)

yln[f(x)] 

解 (1)y f (x2)(x2)2xf (x2)

y2f (x2)2x2xf (x2)2f (x2)4x2f (x2)

(2)y1f(x)

f(x)f(x)f(x)f(x)f(x)f(x)f(x)[f(x)]2

y

22[f(x)][f(x)] 4 试从dx1导出

dyy2y (1)dx

23dy(y)33(y)2yydx (2)3

dy(y)52ddxd1d1dxy1y

 解 (1)dxdy2dydydyydxydy(y)2y(y)33ddydydx

 (2)xdxy3dydy3dyy3y(y)3y3(y)2y13(y)2yy



y(y)6(y)5

5 已知物体的运动规律为sAsint(A、是常数) 求物体运动的加速度 并验证

2ds2s0

2dt 解

dsAcost

dt

2dsA2sint

2dtd2s就是物体运动的加速度

dt22s2sA2sint2Asint0

d2dt 6 验证函数yC1exC2ex(C1

C2是常数)满足关系式

y2y0 

解

yC1exC2ex

yC12exC22ex

y2y(C12exC22ex)2(C1exC2ex)

(C12exC22ex)(C12exC22ex)0 

7 验证函数yexsin

x满足关系式

y2y2y0 

解

yexsin

xexcos

xex(sin

xcos

x)

yex(sin

xcos

x)ex(cos

xsin

x)2excos

x 

y2y2y2excos

x2ex(sin

xcos

x)2exsin

x

2excos

x2exsin

x2excos

x2exsin

x0 

8 求下列函数的n阶导数的一般表达式

(1)

yxna1xn1a2xn2    an1xan

(a1

a2   

an都是常数)

(2)

ysin2x



(3)

yxln

x 

(4)

yxex 

解 (1) ynxn1(n1)a1xn2(n2)a2xn3    an1

yn(n1)xn2(n1)(n2)a1xn3(n2)(n3)a2xn4    an2

  

y(n)n(n1)(n2)  21x0n! 

(2)

y2sin

x cos

xsin2x 

y2cos2x2sin(2x)

2

y22cos(2x)22sin(2x2)

22

y(4)23cos(2x2)23sin(2x3)

22   

y(n)2n1sin[2x(n1)]

2 (3)

ylnx1

y1x1

x

y(1)x2

y(4)(1)(2)x3

  

y(n)(1)(2)(3)  (n2)xn1(1)n2 (4)

yexxex 

yexexxex2exxex 

y2exexxex3exxex 

  

y(n)nexxexex(nx) 

9 求下列函数所指定的阶的导数

(1)

yexcos

x 求y(4) 

(2)

yxsh

x 求y(100)



(3)

yx2sin 2x 求y(50)

.

解 (1)令uex

vcos

x  有

(n2)!n(n2)!

(1)xn1xn1

uuuue

vsin

x 

vcos

x 

vsin

x

v(4)cos

x 

所以

y(4)u(4)v4uv6uv4uvuv(4)

ex[cos

x4(sin

x)6(cos

x)4sin

xcos

x]4excos

x 

(2)令ux

vsh

x 则有

u1

u0

vch

x

vsh

x    

v(99)ch

x 

v(100)sh

x

所以

1298(98)99(99)y(100)u(100)vC100u(99)vC100u(98)v    C100uvC100uvuv(100)

(4)x 100ch

xxsh

x 

(3)令ux2 

vsin 2x 则有

u2x

u2

u0

v(48)248sin(2x48)248sin2x

2

v(49)249cos 2x

v(50)250sin 2x 

12(48)48(48)49(49)u(49)vC50uv    C50uvC50uvuv(50)

所以

y(50)u(50)vC15048(48)49(49)uvC50uvuv(50)

C50

50492228sin2x502x249cos2xx2(250sin2x)

2

250(x2sin2x50xcos2x1225sin2x)

2

2024年2月21日发(作者：聊乐蓉)

高等数学第六版课后习题及答案 第二章第四节

习题 24

1 求函数的二阶导数

(1)

y2x2ln

x

(2)

ye2x1

(3)

yxcos

x

(4)

yet sin

t

(5)ya2x2

(6)

yln(1x2)

(7)

ytan

x

(8)y31

x1 (9)

y(1x2)arctan

x 

xe (10)y

x (11)yxex

(12)yln(x1x2)

解 (1)y4x1

y412

xx (2)

ye2x1

22e2x1

y2e2x1

24e2x1

(3)

yxcos

x



ycos

xxsin

x

ysin

xsin

xxcos

x2sin

xxcos

x 

(4)

yetsin

tetcos

tet(cos

tsin

t)

yet(cos

tsin

t)et(sin

tcos

t)2etcos

t 

2

(5)y1x

(a2x2)22222axaxa2x2xxa2x22

ya2xa2

2222(ax)ax (6)

y12(1x2)2x2

1x1x2(1x2)2x(2x)2(1x2)

y

(1x2)2(1x2)2 (7)

ysec2

x

y2sec

x(sec

x)2sec

xsec

xtan

x2sec2xtan

x 

2(x31)3x (8)y3232

(x1)(x1)6x(x31)23x22(x31)3x6x(2x31)

y

(x31)4(x31)3 (9)y2xarctanx(1x2)122xarctanx1

1x

y2arctanx2x2

1xxxex(x1)exe1 (10)y

x2x2[ex(x1)ex]x2ex(x1)2xex(x22x2)

y

x4x3 (11)yexxex(2x)ex(12x2)

yex2x(12x2)ex4x2xex(32x2)

(12)y11(x1x2)(12x)1

x1x2x1x221x21x2222222x

y12(1x2)122x

1x1x21x2)(1x)21x 2 设f(x)(x10)6

f (2)?

解f (x)6(x10)5

f (x)30(x10)4

f (x)120(x10)3

f (2)120(210)207360

d2y 3 若f (x)存在 求下列函数y的二阶导数2

dx3 (1)

yf(x2)

(2)

yln[f(x)] 

解 (1)y f (x2)(x2)2xf (x2)

y2f (x2)2x2xf (x2)2f (x2)4x2f (x2)

(2)y1f(x)

f(x)f(x)f(x)f(x)f(x)f(x)f(x)[f(x)]2

y

22[f(x)][f(x)] 4 试从dx1导出

dyy2y (1)dx

23dy(y)33(y)2yydx (2)3

dy(y)52ddxd1d1dxy1y

 解 (1)dxdy2dydydyydxydy(y)2y(y)33ddydydx

 (2)xdxy3dydy3dyy3y(y)3y3(y)2y13(y)2yy



y(y)6(y)5

5 已知物体的运动规律为sAsint(A、是常数) 求物体运动的加速度 并验证

2ds2s0

2dt 解

dsAcost

dt

2dsA2sint

2dtd2s就是物体运动的加速度

dt22s2sA2sint2Asint0

d2dt 6 验证函数yC1exC2ex(C1

C2是常数)满足关系式

y2y0 

解

yC1exC2ex

yC12exC22ex

y2y(C12exC22ex)2(C1exC2ex)

(C12exC22ex)(C12exC22ex)0 

7 验证函数yexsin

x满足关系式

y2y2y0 

解

yexsin

xexcos

xex(sin

xcos

x)

yex(sin

xcos

x)ex(cos

xsin

x)2excos

x 

y2y2y2excos

x2ex(sin

xcos

x)2exsin

x

2excos

x2exsin

x2excos

x2exsin

x0 

8 求下列函数的n阶导数的一般表达式

(1)

yxna1xn1a2xn2    an1xan

(a1

a2   

an都是常数)

(2)

ysin2x



(3)

yxln

x 

(4)

yxex 

解 (1) ynxn1(n1)a1xn2(n2)a2xn3    an1

yn(n1)xn2(n1)(n2)a1xn3(n2)(n3)a2xn4    an2

  

y(n)n(n1)(n2)  21x0n! 

(2)

y2sin

x cos

xsin2x 

y2cos2x2sin(2x)

2

y22cos(2x)22sin(2x2)

22

y(4)23cos(2x2)23sin(2x3)

22   

y(n)2n1sin[2x(n1)]

2 (3)

ylnx1

y1x1

x

y(1)x2

y(4)(1)(2)x3

  

y(n)(1)(2)(3)  (n2)xn1(1)n2 (4)

yexxex 

yexexxex2exxex 

y2exexxex3exxex 

  

y(n)nexxexex(nx) 

9 求下列函数所指定的阶的导数

(1)

yexcos

x 求y(4) 

(2)

yxsh

x 求y(100)



(3)

yx2sin 2x 求y(50)

.

解 (1)令uex

vcos

x  有

(n2)!n(n2)!

(1)xn1xn1

uuuue

vsin

x 

vcos

x 

vsin

x

v(4)cos

x 

所以

y(4)u(4)v4uv6uv4uvuv(4)

ex[cos

x4(sin

x)6(cos

x)4sin

xcos

x]4excos

x 

(2)令ux

vsh

x 则有

u1

u0

vch

x

vsh

x    

v(99)ch

x 

v(100)sh

x

所以

1298(98)99(99)y(100)u(100)vC100u(99)vC100u(98)v    C100uvC100uvuv(100)

(4)x 100ch

xxsh

x 

(3)令ux2 

vsin 2x 则有

u2x

u2

u0

v(48)248sin(2x48)248sin2x

2

v(49)249cos 2x

v(50)250sin 2x 

12(48)48(48)49(49)u(49)vC50uv    C50uvC50uvuv(50)

所以

y(50)u(50)vC15048(48)49(49)uvC50uvuv(50)

C50

50492228sin2x502x249cos2xx2(250sin2x)

2

250(x2sin2x50xcos2x1225sin2x)

2