第二章一元函数微分学(3)

2019-06-11 00:20

y??(?r?x)???xr?x22

22(r?x)?2r?x22

22

??22解法二:

(x?y)??(r)?2x?2y?y??0

2

y???

xy??xr?x22

y?cos(xy) ⑸

??[cos(xy)]???sin(xy)?(xy)?y解法一:

??sin(xy)?(y?xy?)

y??∴

解法二:设

?ysin(xy)1?xsin(xy)

F(x,y)?y?cos(xy)

Fy??1?xsin(xy)

Fx??ysin(xy),dy

dx??Fx?Fy???ysinxy()1?xsinxy()

ylnx?xlny ⑹

??(xlny)?(ylnx)解法一:

lnx?y??yx?lny?yxxyxyy?

y??lny?lnx??xylny?yxylnx?x22

解法二:设

F(x,y)?ylnx?xlny

Fx??yx?lny,Fy??lnx?yxxyxy

dydx

??Fx?Fy????lny?lnx?xylny?yxylnx?x22y? ⑺

(x?1)(x?2)x?3

解:(对数法)

lny?ln?12(x?1)(x?2)x?3

[ln(x?1)?ln(x?2)?ln(x?3)]

(lny)??{1[ln(x?1)?ln(x?2)?ln(x?3)]}?2

1y ∴

y??12x?11(1?1x?2?1x?3)

y??

12x?1x

(?1x?2?1x?3)(x?1)(x?2)x?3

y?x解法一:(对数法)

lny?lnx?xlnx1y∴

x

y??lnx?xxx?lnx?1

y??x(lnx?1)

解法二:(指数法)

y?x?exlnxx?exlnx

y??(e

xlnx)??exlnx(xlnx)?

?x(lnx?1)xx

y?2x ⑼

解法一:(对数法)

?(sinx)xcosx

y?2x1设

,y2?(sinx)cosx

y?y1?y2,lny1?ln2x1y??12xx??y2?y??y1x

?ln2?xx?xlnx

12xx?12

ylnx?(lnx?2)

y??∴

2x2x(lnx?2)?x(lnx?2)

lny2?cosxlnsinx1y2

???sinxlnsinx?cosxy2cosxsinx

??(sinx)y2cosx(cosxcotx?sinxlnsinx)

??y2?y??y1?xx?12

(lnx?1)?(sinx)xlnxcosx(cosxcotx?sinxlnsinx解法二:(指数法)

y?2e?2e

?ecosxlnsinx

xlnxcosxlnsinx?(xlnx)?e(cosxlnsinx)??x

x?12(lnx?1)?(sinx)y

cosx(cosxcotx?sinxlnsin ⑽

y?xxxlny?ylnx解法一:

lny?xyy??lnx?y??xylny?yxylnx?xyx

yx

22

y??∴

解法二:设

F(x,y)?x?yy?1Fx??yx

?ylny?xyxx?ylny?(yxyx?lny)xy


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